\displaystyle \textbf{Question 1: }\text{Which of the following cannot be valid assignments of probability}
\displaystyle \text{for the elementary outcomes of the sample space}
\displaystyle S=\{w_1,w_2,w_3,w_4,w_5,w_6,w_7\}?

\displaystyle \begin{array} {c | c c c c c c c} \text{Elementary outcomes} & w_1 & w_2 & w_3 & w_4 & w_5 & w_6 & w_7  \\ \hline \text{(i)} & 0.1 & 0.01 & 0.05 & 0.03 & 0.01 & 0.2 & 0.6 \\ \hline \text{(ii)} & \frac{1}{7} & \frac{1}{7} & \frac{1}{7} & \frac{1}{7} & \frac{1}{7} & \frac{1}{7} & \frac{1}{7} \\ \hline \text{(iii)} & 0.7 & 0.6 & 0.5 & 0.4 & 0.3 & 0.2 & 0.1 \\ \hline \text{(iv)} & \frac{1}{14} & \frac{2}{14} & \frac{3}{14} & \frac{4}{14} & \frac{5}{14} & \frac{6}{14} & \frac{15}{14} \\ \hline \end{array}

\displaystyle \textbf{Answer:}
\displaystyle \text{For a valid probability assignment, }0\leq P(w_i)\leq1
\displaystyle \text{for each }i,\text{ and }\sum_{i=1}^{7}P(w_i)=1.
\displaystyle \text{(i) }\sum_{i=1}^{7}P(w_i)=0.1+0.01+0.05+0.03+0.01+0.2+0.6=1.
\displaystyle \text{Also, each probability lies between }0\text{ and }1.
\displaystyle \therefore \text{Assignment (i) is valid.}
\displaystyle \text{(ii) }\sum_{i=1}^{7}P(w_i)=7\times\frac17=1.
\displaystyle \text{Also, each probability lies between }0\text{ and }1.
\displaystyle \therefore \text{Assignment (ii) is valid.}
\displaystyle \text{(iii) }\sum_{i=1}^{7}P(w_i)=0.7+0.6+0.5+0.4+0.3+0.2+0.1=2.8.
\displaystyle \text{Since the sum of the probabilities is not }1,\text{ assignment (iii) is invalid.}
\displaystyle \text{(iv) }P(w_7)=\frac{15}{14}>1.
\displaystyle \text{Therefore, assignment (iv) is invalid.}
\displaystyle \therefore \text{Assignments (iii) and (iv) cannot be valid probability assignments.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{A die is thrown. Find the probability of getting:}
\displaystyle \text{(i) a prime number}\qquad\text{(ii) }2\text{ or }4\qquad\text{(iii) a multiple of }2\text{ or }3
\displaystyle \text{(iv) an even prime number}\qquad\text{(v) a number greater than }5
\displaystyle \text{(vi) a number lying between }2\text{ and }6
\displaystyle \textbf{Answer:}
\displaystyle \text{The sample space is }S=\{1,2,3,4,5,6\}.
\displaystyle \text{(i) Prime numbers are }2,3\text{ and }5.
\displaystyle P(\text{prime number})=\frac{3}{6}=\frac12.
\displaystyle \text{(ii) The favourable outcomes are }2\text{ and }4.
\displaystyle P(2\text{ or }4)=\frac{2}{6}=\frac13.
\displaystyle \text{(iii) Multiples of }2\text{ or }3\text{ are }2,3,4\text{ and }6.
\displaystyle P(\text{multiple of }2\text{ or }3)=\frac{4}{6}=\frac23.
\displaystyle \text{(iv) The only even prime number is }2.
\displaystyle P(\text{even prime number})=\frac16.
\displaystyle \text{(v) The only number greater than }5\text{ is }6.
\displaystyle P(\text{number greater than }5)=\frac16.
\displaystyle \text{(vi) The numbers lying between }2\text{ and }6\text{ are }3,4\text{ and }5.
\displaystyle P(\text{number lying between }2\text{ and }6)=\frac36=\frac12.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{In a simultaneous throw of a pair of dice, find the}
\displaystyle \text{probability of getting:}
\displaystyle \text{(i) }8\text{ as the sum}\qquad\text{(ii) a doublet}\qquad\text{(iii) a doublet of prime numbers}
\displaystyle \text{(iv) a doublet of odd numbers}\qquad\text{(v) a sum greater than }9
\displaystyle \text{(vi) an even number on the first die}
\displaystyle \text{(vii) an even number on one die and a multiple of }3\text{ on the other}
\displaystyle \text{(viii) neither }9\text{ nor }11\text{ as the sum of the numbers on the faces}
\displaystyle \text{(ix) a sum less than }6\qquad\text{(x) a sum less than }7
\displaystyle \text{(xi) a sum more than }7\qquad\text{(xii) neither a doublet nor a total of }10
\displaystyle \text{(xiii) an odd number on the first die and }6\text{ on the second die}
\displaystyle \text{(xiv) a number greater than }4\text{ on each die}\qquad\text{(xv) a total of }9\text{ or }11
\displaystyle \text{(xvi) a total greater than }8
\displaystyle \textbf{Answer:}
\displaystyle \text{When two dice are thrown, the total number of equally likely outcomes}=6^2=36.
\displaystyle \text{(i) The outcomes having sum }8\text{ are}
\displaystyle \{(2,6),(3,5),(4,4),(5,3),(6,2)\}.
\displaystyle \therefore P(\text{sum }8)=\frac{5}{36}.
\displaystyle \text{(ii) The doublets are}
\displaystyle \{(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)\}.
\displaystyle \therefore P(\text{doublet})=\frac{6}{36}=\frac16.
\displaystyle \text{(iii) The doublets of prime numbers are }
\displaystyle \{(2,2),(3,3),(5,5)\}.
\displaystyle \therefore P(\text{doublet of prime numbers})=\frac{3}{36}=\frac{1}{12}.
\displaystyle \text{(iv) The doublets of odd numbers are }
\displaystyle \{(1,1),(3,3),(5,5)\}.
\displaystyle \therefore P(\text{doublet of odd numbers})=\frac{3}{36}=\frac{1}{12}.
\displaystyle \text{(v) The outcomes having sum greater than }9\text{ are}
\displaystyle \{(4,6),(5,5),(5,6),(6,4),(6,5),(6,6)\}.
\displaystyle \therefore P(\text{sum greater than }9)=\frac{6}{36}=\frac16.
\displaystyle \text{(vi) The first die may show }2,4\text{ or }6,
\displaystyle \text{while the second die may show any of the six numbers.}
\displaystyle \therefore \text{Number of favourable outcomes}=3\times6=18.
\displaystyle P(\text{even number on the first die})=\frac{18}{36}=\frac12.
\displaystyle \text{(vii) The favourable outcomes are}
\displaystyle \{(2,3),(2,6),(4,3),(4,6),(6,3),(6,6),
\displaystyle \phantom{\{}(3,2),(3,4),(3,6),(6,2),(6,4)\}.
\displaystyle \therefore P(\text{required event})=\frac{11}{36}.
\displaystyle \text{(viii) The outcomes having sum }9\text{ or }11\text{ are}
\displaystyle \{(3,6),(4,5),(5,4),(6,3),(5,6),(6,5)\}.
\displaystyle P(\text{sum }9\text{ or }11)=\frac{6}{36}=\frac16.
\displaystyle \therefore P(\text{neither sum }9\text{ nor }11)=1-\frac16=\frac56.
\displaystyle \text{(ix) The outcomes having sum less than }6\text{ are}
\displaystyle \{(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),
\displaystyle \phantom{\{}(2,3),(3,1),(3,2),(4,1)\}.
\displaystyle \therefore P(\text{sum less than }6)=\frac{10}{36}=\frac{5}{18}.
\displaystyle \text{(x) The possible sums less than }7\text{ are }2,3,4,5\text{ and }6.
\displaystyle \text{The corresponding numbers of outcomes are }1,2,3,4\text{ and }5.
\displaystyle \therefore P(\text{sum less than }7)=\frac{1+2+3+4+5}{36}
\displaystyle =\frac{15}{36}=\frac{5}{12}.
\displaystyle \text{(xi) The possible sums greater than }7\text{ are }8,9,10,11\text{ and }12.
\displaystyle \text{The corresponding numbers of outcomes are }5,4,3,2\text{ and }1.
\displaystyle \therefore P(\text{sum more than }7)=\frac{5+4+3+2+1}{36}
\displaystyle =\frac{15}{36}=\frac{5}{12}.
\displaystyle \text{(xii) Let }A\text{ be the event of getting a doublet and }B\text{ the event}
\displaystyle \text{of getting a total of }10.
\displaystyle A=\{(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)\}.
\displaystyle B=\{(4,6),(5,5),(6,4)\}.
\displaystyle A\cap B=\{(5,5)\}.
\displaystyle n(A\cup B)=6+3-1=8.
\displaystyle \therefore \text{Number of outcomes which are neither a doublet nor a total of }10
\displaystyle =36-8=28.
\displaystyle P(\text{neither a doublet nor a total of }10)=\frac{28}{36}=\frac79.
\displaystyle \text{(xiii) The favourable outcomes are }
\displaystyle \{(1,6),(3,6),(5,6)\}.
\displaystyle \therefore P(\text{odd on the first die and }6\text{ on the second})
\displaystyle =\frac{3}{36}=\frac{1}{12}.
\displaystyle \text{(xiv) A number greater than }4\text{ on each die means that each die}
\displaystyle \text{shows either }5\text{ or }6.
\displaystyle \text{The favourable outcomes are }\{(5,5),(5,6),(6,5),(6,6)\}.
\displaystyle \therefore P(\text{number greater than }4\text{ on each die})
\displaystyle =\frac{4}{36}=\frac19.
\displaystyle \text{(xv) The outcomes having a total of }9\text{ or }11\text{ are}
\displaystyle \{(3,6),(4,5),(5,4),(6,3),(5,6),(6,5)\}.
\displaystyle \therefore P(\text{total }9\text{ or }11)=\frac{6}{36}=\frac16.
\displaystyle \text{(xvi) The possible totals greater than }8\text{ are }9,10,11\text{ and }12.
\displaystyle \text{The corresponding numbers of outcomes are }4,3,2\text{ and }1.
\displaystyle \therefore P(\text{total greater than }8)=\frac{4+3+2+1}{36}
\displaystyle =\frac{10}{36}=\frac{5}{18}.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{In a single throw of three dice, find the probability of}
\displaystyle \text{getting a total of }17\text{ or }18.
\displaystyle \textbf{Answer:}
\displaystyle \text{When three dice are thrown, the total number of possible outcomes is}
\displaystyle 6^3=216.
\displaystyle \therefore n(S)=216.
\displaystyle \text{Let }E\text{ be the event of getting a total of }17\text{ or }18.
\displaystyle E=\{(6,6,5),(6,5,6),(5,6,6),(6,6,6)\}.
\displaystyle \therefore n(E)=4.
\displaystyle P(E)=\frac{n(E)}{n(S)}=\frac{4}{216}=\frac{1}{54}.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Three coins are tossed together. Find the probability of}
\displaystyle \text{getting: (i) exactly two heads (ii) at least two heads}
\displaystyle \text{(iii) at least one head and one tail.}
\displaystyle \textbf{Answer:}
\displaystyle \text{When three coins are tossed, the total number of possible outcomes is}
\displaystyle 2^3=8.
\displaystyle \therefore n(S)=8.
\displaystyle S=\{HHH,HHT,HTH,THH,HTT,THT,TTH,TTT\}.
\displaystyle \text{(i) Exactly two heads: }E=\{HHT,HTH,THH\}.
\displaystyle \therefore n(E)=3,\qquad P(E)=\frac{3}{8}.
\displaystyle \text{(ii) At least two heads: }E=\{HHH,HHT,HTH,THH\}.
\displaystyle \therefore n(E)=4,\qquad P(E)=\frac{4}{8}=\frac12.
\displaystyle \text{(iii) At least one head and one tail:}
\displaystyle E=\{HHT,HTH,THH,HTT,THT,TTH\}.
\displaystyle \therefore n(E)=6,\qquad P(E)=\frac{6}{8}=\frac34.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{What is the probability that an ordinary year has}
\displaystyle \text{53 Sundays?}
\displaystyle \textbf{Answer:}
\displaystyle \text{An ordinary year has }365\text{ days.}
\displaystyle 365=52\times7+1.
\displaystyle \text{Thus, an ordinary year has }52\text{ complete weeks and }1\text{ extra day.}
\displaystyle \text{The extra day may be any one of the seven days of the week.}
\displaystyle \therefore n(S)=7.
\displaystyle \text{The year has }53\text{ Sundays only when the extra day is Sunday.}
\displaystyle \therefore n(E)=1.
\displaystyle P(E)=\frac{n(E)}{n(S)}=\frac17.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{What is the probability that a leap year has}
\displaystyle \text{53 Sundays and 53 Mondays?}
\displaystyle \textbf{Answer:}
\displaystyle \text{A leap year has }366\text{ days.}
\displaystyle 366=52\times7+2.
\displaystyle \text{Thus, a leap year has }52\text{ complete weeks and }2\text{ extra days.}
\displaystyle \text{The possible pairs of extra days are}
\displaystyle (\text{Sunday, Monday}),(\text{Monday, Tuesday}),
\displaystyle (\text{Tuesday, Wednesday}),(\text{Wednesday, Thursday}),
\displaystyle (\text{Thursday, Friday}),(\text{Friday, Saturday}),
\displaystyle (\text{Saturday, Sunday}).
\displaystyle \therefore n(S)=7.
\displaystyle \text{For the year to have }53\text{ Sundays and }53\text{ Mondays,}
\displaystyle \text{the two extra days must be Sunday and Monday.}
\displaystyle \therefore n(E)=1.
\displaystyle P(E)=\frac{n(E)}{n(S)}=\frac17.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{A bag contains }8\text{ red and }5\text{ white balls. Three balls are}
\displaystyle \text{drawn at random. Find the probability that:}
\displaystyle \text{(i) All the three balls are white}\qquad\text{(ii) All the three balls are red}
\displaystyle \text{(iii) One ball is red and two balls are white}
\displaystyle \textbf{Answer:}
\displaystyle \text{Total number of balls}=13.
\displaystyle \text{Hence, the total number of possible outcomes}=\,^{13}C_3.
\displaystyle \text{(i) Number of favourable outcomes}=\,^{5}C_3.
\displaystyle P(\text{all three white})=\frac{\,^{5}C_3}{\,^{13}C_3}
\displaystyle =\frac{10}{286}=\frac{5}{143}.
\displaystyle \text{(ii) Number of favourable outcomes}=\,^{8}C_3.
\displaystyle P(\text{all three red})=\frac{\,^{8}C_3}{\,^{13}C_3}
\displaystyle =\frac{56}{286}=\frac{28}{143}.
\displaystyle \text{(iii) Number of favourable outcomes}=\,^{8}C_1\times\,^{5}C_2.
\displaystyle P(\text{one red and two white})=\frac{\,^{8}C_1\times\,^{5}C_2}{\,^{13}C_3}
\displaystyle =\frac{80}{286}=\frac{40}{143}.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{In a single throw of three dice, find the probability of}
\displaystyle \text{getting the same number on all the three dice.}
\displaystyle \textbf{Answer:}
\displaystyle \text{When three dice are thrown, the total number of possible outcomes is}
\displaystyle 6^3=216.
\displaystyle \therefore n(S)=216.
\displaystyle \text{Let }E\text{ be the event of getting the same number on all three dice.}
\displaystyle E=\{(1,1,1),(2,2,2),(3,3,3),(4,4,4),(5,5,5),(6,6,6)\}.
\displaystyle \therefore n(E)=6.
\displaystyle P(E)=\frac{n(E)}{n(S)}=\frac{6}{216}=\frac{1}{36}.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Two unbiased dice are thrown. Find the probability that}
\displaystyle \text{the total of the numbers on the dice is greater than }10.
\displaystyle \textbf{Answer:}
\displaystyle \text{When two dice are thrown, the total number of possible outcomes is}
\displaystyle 6^2=36.
\displaystyle \therefore n(S)=36.
\displaystyle \text{Let }E\text{ be the event of getting a sum greater than }10.
\displaystyle E=\{(5,6),(6,5),(6,6)\}.
\displaystyle \therefore n(E)=3.
\displaystyle P(E)=\frac{n(E)}{n(S)}=\frac{3}{36}=\frac{1}{12}.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{A card is drawn at random from a pack of }52\text{ cards.}
\displaystyle \text{Find the probability that the card drawn is:}
\displaystyle \text{(i) a black king}\qquad\text{(ii) either a black card or a king}
\displaystyle \text{(iii) black and a king}\qquad\text{(iv) a jack, queen or king}
\displaystyle \text{(v) neither a heart nor a king}\qquad\text{(vi) a spade or an ace}
\displaystyle \text{(vii) neither an ace nor a king}\qquad\text{(viii) neither a red card nor a queen}
\displaystyle \text{(ix) other than an ace}\qquad\text{(x) a ten}\qquad\text{(xi) a spade}
\displaystyle \text{(xii) a black card}\qquad\text{(xiii) the seven of clubs}\qquad\text{(xiv) a jack}
\displaystyle \text{(xv) the ace of spades}\qquad\text{(xvi) a queen}\qquad\text{(xvii) a heart}
\displaystyle \text{(xviii) a red card}
\displaystyle \textbf{Answer:}
\displaystyle \text{The total number of cards in a standard pack is }52.
\displaystyle \therefore n(S)=52.
\displaystyle \text{(i) There are }2\text{ black kings: the King of Spades and the King of Clubs.}
\displaystyle P(\text{black king})=\frac{2}{52}=\frac{1}{26}.
\displaystyle \text{(ii) There are }26\text{ black cards and }4\text{ kings, of which }2\text{ are black.}
\displaystyle n(E)=26+4-2=28.
\displaystyle P(\text{black card or king})=\frac{28}{52}=\frac{7}{13}.
\displaystyle \text{(iii) A card which is black and a king may be the King of Spades}
\displaystyle \text{or the King of Clubs.}
\displaystyle P(\text{black and king})=\frac{2}{52}=\frac{1}{26}.
\displaystyle \text{(iv) There are }4\text{ jacks, }4\text{ queens and }4\text{ kings.}
\displaystyle n(E)=4+4+4=12.
\displaystyle P(\text{jack, queen or king})=\frac{12}{52}=\frac{3}{13}.
\displaystyle \text{(v) There are }13\text{ hearts and }3\text{ kings which are not hearts.}
\displaystyle n(E)=52-13-3=36.
\displaystyle P(\text{neither heart nor king})=\frac{36}{52}=\frac{9}{13}.
\displaystyle \text{(vi) There are }13\text{ spades and }3\text{ aces which are not spades.}
\displaystyle n(E)=13+3=16.
\displaystyle P(\text{spade or ace})=\frac{16}{52}=\frac{4}{13}.
\displaystyle \text{(vii) There are }4\text{ aces and }4\text{ kings.}
\displaystyle n(E)=52-4-4=44.
\displaystyle P(\text{neither ace nor king})=\frac{44}{52}=\frac{11}{13}.
\displaystyle \text{(viii) There are }26\text{ black cards, of which }2\text{ are queens.}
\displaystyle n(E)=26-2=24.
\displaystyle P(\text{neither red nor queen})=\frac{24}{52}=\frac{6}{13}.
\displaystyle \text{(ix) There are }4\text{ aces in the pack.}
\displaystyle n(E)=52-4=48.
\displaystyle P(\text{other than an ace})=\frac{48}{52}=\frac{12}{13}.
\displaystyle \text{(x) There are }4\text{ tens in the pack.}
\displaystyle P(\text{ten})=\frac{4}{52}=\frac{1}{13}.
\displaystyle \text{(xi) There are }13\text{ spades in the pack.}
\displaystyle P(\text{spade})=\frac{13}{52}=\frac14.
\displaystyle \text{(xii) There are }26\text{ black cards in the pack.}
\displaystyle P(\text{black card})=\frac{26}{52}=\frac12.
\displaystyle \text{(xiii) There is only one seven of clubs.}
\displaystyle P(\text{seven of clubs})=\frac{1}{52}.
\displaystyle \text{(xiv) There are }4\text{ jacks in the pack.}
\displaystyle P(\text{jack})=\frac{4}{52}=\frac{1}{13}.
\displaystyle \text{(xv) There is only one ace of spades.}
\displaystyle P(\text{ace of spades})=\frac{1}{52}.
\displaystyle \text{(xvi) There are }4\text{ queens in the pack.}
\displaystyle P(\text{queen})=\frac{4}{52}=\frac{1}{13}.
\displaystyle \text{(xvii) There are }13\text{ hearts in the pack.}
\displaystyle P(\text{heart})=\frac{13}{52}=\frac14.
\displaystyle \text{(xviii) There are }26\text{ red cards in the pack.}
\displaystyle P(\text{red card})=\frac{26}{52}=\frac12.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{In shuffling a pack of }52\text{ playing cards, four cards are}
\displaystyle \text{accidentally dropped. Find the probability that the missing cards}
\displaystyle \text{are one from each suit.}
\displaystyle \textbf{Answer:}
\displaystyle \text{The total number of ways of selecting }4\text{ missing cards is}
\displaystyle n(S)=\,^{52}C_4=270725.
\displaystyle \text{For one card to be missing from each suit, one card must be selected}
\displaystyle \text{from each of the four suits.}
\displaystyle n(E)=\,^{13}C_1\times\,^{13}C_1\times\,^{13}C_1\times\,^{13}C_1=13^4.
\displaystyle P(E)=\frac{n(E)}{n(S)}=\frac{13^4}{270725}
\displaystyle =\frac{28561}{270725}=\frac{2197}{20825}.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{From a deck of }52\text{ cards, four cards are drawn}
\displaystyle \text{simultaneously. Find the probability that they are the four honours}
\displaystyle \text{of the same suit.}
\displaystyle \textbf{Answer:}
\displaystyle \text{The four honours of a suit are Ace, King, Queen and Jack.}
\displaystyle \text{The total number of ways of drawing }4\text{ cards is}
\displaystyle n(S)=\,^{52}C_4=270725.
\displaystyle \text{There is one favourable set of four honours for each of the four suits.}
\displaystyle \therefore n(E)=4\times\,^{4}C_4=4.
\displaystyle P(E)=\frac{n(E)}{n(S)}=\frac{4}{270725}.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{Tickets numbered from }1\text{ to }20\text{ are mixed together and}
\displaystyle \text{one ticket is drawn at random. Find the probability that its number}
\displaystyle \text{is a multiple of }3\text{ or }7.
\displaystyle \textbf{Answer:}
\displaystyle \text{The total number of possible outcomes is }n(S)=20.
\displaystyle \text{The multiples of }3\text{ or }7\text{ from }1\text{ to }20\text{ are}
\displaystyle E=\{3,6,7,9,12,14,15,18\}.
\displaystyle \therefore n(E)=8.
\displaystyle P(E)=\frac{n(E)}{n(S)}=\frac{8}{20}=\frac25.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{A bag contains }6\text{ red, }4\text{ white and }8\text{ blue balls.}
\displaystyle \text{If three balls are drawn at random, find the probability that one}
\displaystyle \text{is red, one is white and one is blue.}
\displaystyle \textbf{Answer:}
\displaystyle \text{Total number of balls}=6+4+8=18.
\displaystyle \text{The total number of ways of drawing }3\text{ balls is}
\displaystyle n(S)=\,^{18}C_3=816.
\displaystyle \text{Let }E\text{ be the event that one ball of each colour is drawn.}
\displaystyle n(E)=\,^{6}C_1\times\,^{4}C_1\times\,^{8}C_1
\displaystyle =6\times4\times8=192.
\displaystyle P(E)=\frac{n(E)}{n(S)}=\frac{192}{816}=\frac{4}{17}.
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{A bag contains }7\text{ white, }5\text{ black and }4\text{ red balls.}
\displaystyle \text{If two balls are drawn at random, find the probability that:}
\displaystyle \text{(i) both balls are white}\qquad\text{(ii) one ball is black and the other is red}
\displaystyle \text{(iii) both balls are of the same colour.}
\displaystyle \textbf{Answer:}
\displaystyle \text{Total number of balls}=7+5+4=16.
\displaystyle \text{The total number of ways of drawing }2\text{ balls is}
\displaystyle n(S)=\,^{16}C_2=120.
\displaystyle \text{(i) Let }E\text{ be the event that both balls are white.}
\displaystyle n(E)=\,^{7}C_2=21.
\displaystyle P(E)=\frac{n(E)}{n(S)}=\frac{21}{120}=\frac{7}{40}.
\displaystyle \text{(ii) Let }E\text{ be the event that one ball is black and the other is red.}
\displaystyle n(E)=\,^{5}C_1\times\,^{4}C_1=5\times4=20.
\displaystyle P(E)=\frac{n(E)}{n(S)}=\frac{20}{120}=\frac16.
\displaystyle \text{(iii) Both balls may be white, black or red.}
\displaystyle n(E)=\,^{7}C_2+\,^{5}C_2+\,^{4}C_2
\displaystyle =21+10+6=37.
\displaystyle P(E)=\frac{n(E)}{n(S)}=\frac{37}{120}.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{A bag contains }6\text{ red, }4\text{ white and }8\text{ blue balls.}
\displaystyle \text{If three balls are drawn at random, find the probability that:}
\displaystyle \text{(i) one is red and two are white}\qquad\text{(ii) two are blue and one is red}
\displaystyle \text{(iii) exactly one is red.}
\displaystyle \textbf{Answer:}
\displaystyle \text{Total number of balls}=6+4+8=18.
\displaystyle \text{The total number of ways of drawing }3\text{ balls is}
\displaystyle n(S)=\,^{18}C_3=816.
\displaystyle \text{(i) Let }E\text{ be the event that one ball is red and two are white.}
\displaystyle n(E)=\,^{6}C_1\times\,^{4}C_2=6\times6=36.
\displaystyle P(E)=\frac{n(E)}{n(S)}=\frac{36}{816}=\frac{3}{68}.
\displaystyle \text{(ii) Let }E\text{ be the event that two balls are blue and one is red.}
\displaystyle n(E)=\,^{8}C_2\times\,^{6}C_1=28\times6=168.
\displaystyle P(E)=\frac{n(E)}{n(S)}=\frac{168}{816}=\frac{7}{34}.
\displaystyle \text{(iii) For exactly one red ball, the other two balls must be selected}
\displaystyle \text{from the }4+8=12\text{ non-red balls.}
\displaystyle n(E)=\,^{6}C_1\times\,^{12}C_2=6\times66=396.
\displaystyle P(E)=\frac{n(E)}{n(S)}=\frac{396}{816}=\frac{33}{68}.
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{Five cards are drawn from a pack of }52\text{ cards. What is the}
\displaystyle \text{probability that these }5\text{ cards contain: (i) exactly one ace}
\displaystyle \text{(ii) at least one ace?}
\displaystyle \textbf{Answer:}
\displaystyle \text{The total number of ways of drawing }5\text{ cards is}
\displaystyle n(S)=\,^{52}C_5=2598960.
\displaystyle \text{(i) For exactly one ace, choose }1\text{ of the }4\text{ aces and }4\text{ of the}
\displaystyle 48\text{ non-ace cards.}
\displaystyle n(E)=\,^{4}C_1\times\,^{48}C_4=778320.
\displaystyle P(E)=\frac{n(E)}{n(S)}=\frac{778320}{2598960}
\displaystyle =\frac{3243}{10829}.
\displaystyle \text{(ii) The event of getting at least one ace is the complement of}
\displaystyle \text{getting no ace.}
\displaystyle P(\text{at least one ace})=1-P(\text{no ace})
\displaystyle =1-\frac{\,^{48}C_5}{\,^{52}C_5}
\displaystyle =1-\frac{1712304}{2598960}
\displaystyle =\frac{886656}{2598960}=\frac{18472}{54145}.
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{The face cards are removed from a full pack. Out of the}
\displaystyle \text{remaining }40\text{ cards, }4\text{ are drawn at random. Find the probability}
\displaystyle \text{that they belong to different suits.}
\displaystyle \textbf{Answer:}
\displaystyle \text{After removing the face cards, }40\text{ cards remain, with }10\text{ cards}
\displaystyle \text{in each of the four suits.}
\displaystyle \text{The total number of ways of drawing }4\text{ cards is}
\displaystyle n(S)=\,^{40}C_4=91390.
\displaystyle \text{For the cards to belong to different suits, one card must be chosen}
\displaystyle \text{from each suit.}
\displaystyle n(E)=\,^{10}C_1\times\,^{10}C_1\times\,^{10}C_1\times\,^{10}C_1
\displaystyle =10^4=10000.
\displaystyle P(E)=\frac{n(E)}{n(S)}=\frac{10000}{91390}=\frac{1000}{9139}.
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{There are four men and six women on the city council. If one council member is}
\displaystyle \text{selected at random for a committee, what is the probability that the selected member is a woman?}
\displaystyle \text{Answer:}
\displaystyle \text{There are }4\text{ men and }6\text{ women on the city council.}
\displaystyle \text{Total number of council members}=4+6=10.
\displaystyle \therefore n(S)={}^{10}C_1=10.
\displaystyle \text{Let }E\text{ be the event that the selected member is a woman.}
\displaystyle n(E)={}^{6}C_1=6.
\displaystyle P(E)=\frac{n(E)}{n(S)}=\frac{6}{10}=\frac{3}{5}.
\displaystyle \therefore \text{The probability that the selected member is a woman is }\frac{3}{5}.
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{A box contains }100\text{ bulbs, }20\text{ of which are defective. Ten bulbs are selected}
\displaystyle \text{for inspection. Find the probability that:}
\displaystyle \text{(i) all }10\text{ are defective}\qquad\text{(ii) all }10\text{ are good}
\displaystyle \text{(iii) at least one is defective}\qquad\text{(iv) none is defective.}
\displaystyle \text{Answer:}
\displaystyle \text{Number of defective bulbs}=20.
\displaystyle \text{Number of good bulbs}=100-20=80.
\displaystyle \text{Total number of ways of selecting }10\text{ bulbs from }100={}^{100}C_{10}.
\displaystyle \therefore n(S)={}^{100}C_{10}.
\displaystyle \text{(i) Let }E_1\text{ be the event that all }10\text{ selected bulbs are defective.}
\displaystyle n(E_1)={}^{20}C_{10}.
\displaystyle P(E_1)=\frac{n(E_1)}{n(S)}=\frac{{}^{20}C_{10}}{{}^{100}C_{10}}.
\displaystyle \text{(ii) Let }E_2\text{ be the event that all }10\text{ selected bulbs are good.}
\displaystyle n(E_2)={}^{80}C_{10}.
\displaystyle P(E_2)=\frac{n(E_2)}{n(S)}=\frac{{}^{80}C_{10}}{{}^{100}C_{10}}.
\displaystyle \text{(iii) Let }E_3\text{ be the event that at least one selected bulb is defective.}
\displaystyle \text{Then }E_3'\text{ is the event that none of the selected bulbs is defective.}
\displaystyle n(E_3')={}^{80}C_{10}.
\displaystyle P(E_3')=\frac{n(E_3')}{n(S)}=\frac{{}^{80}C_{10}}{{}^{100}C_{10}}.
\displaystyle \therefore P(E_3)=1-P(E_3')=1-\frac{{}^{80}C_{10}}{{}^{100}C_{10}}.
\displaystyle \text{(iv) Let }E_4\text{ be the event that none of the selected bulbs is defective.}
\displaystyle \text{Thus, all }10\text{ selected bulbs must be good.}
\displaystyle n(E_4)={}^{80}C_{10}.
\displaystyle P(E_4)=\frac{n(E_4)}{n(S)}=\frac{{}^{80}C_{10}}{{}^{100}C_{10}}.
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{Find the probability that in a random arrangement of the letters of the word}
\displaystyle \text{`SOCIAL', the vowels come together.}
\displaystyle \text{Answer:}
\displaystyle \text{The word SOCIAL has }6\text{ distinct letters.}
\displaystyle \text{Total number of arrangements}=6!.
\displaystyle \therefore n(S)=6!.
\displaystyle \text{Let }E\text{ be the event that all the vowels come together.}
\displaystyle \text{The vowels are }A,\ I,\ O.
\displaystyle \text{Treat the three vowels as one block. Then we have }4\text{ objects: }(AIO),S,C,L.
\displaystyle \text{These }4\text{ objects can be arranged in }4!\text{ ways.}
\displaystyle \text{The vowels within the block can be arranged in }3!\text{ ways.}
\displaystyle \therefore n(E)=4!\times3!.
\displaystyle P(E)=\frac{n(E)}{n(S)}=\frac{4!\times3!}{6!}=\frac{1}{5}.
\displaystyle \therefore \text{The required probability is }\frac{1}{5}.
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{The letters of the word `CLIFTON' are placed at random in a row. What is the}
\displaystyle \text{probability that the two vowels come together?}
\displaystyle \text{Answer:}
\displaystyle \text{The word CLIFTON has }7\text{ distinct letters.}
\displaystyle \text{Total number of arrangements}=7!.
\displaystyle \therefore n(S)=7!.
\displaystyle \text{Let }E\text{ be the event that the two vowels come together.}
\displaystyle \text{The vowels are }I\text{ and }O.
\displaystyle \text{Treat the two vowels as one block. Then we have }6\text{ objects: }(IO),C,L,F,T,N.
\displaystyle \text{These }6\text{ objects can be arranged in }6!\text{ ways.}
\displaystyle \text{The vowels within the block can be arranged in }2!\text{ ways.}
\displaystyle \therefore n(E)=6!\times2!.
\displaystyle P(E)=\frac{n(E)}{n(S)}=\frac{6!\times2!}{7!}=\frac{2}{7}.
\displaystyle \therefore \text{The required probability is }\frac{2}{7}.
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{The letters of the word `FORTUNATES' are arranged at random in a row. What is the}
\displaystyle \text{probability that the two }T\text{'s come together?}
\displaystyle \text{Answer:}
\displaystyle \text{The word FORTUNATES has }10\text{ letters, of which }T\text{ occurs twice.}
\displaystyle \text{Total number of distinct arrangements}=\frac{10!}{2!}.
\displaystyle \therefore n(S)=\frac{10!}{2!}.
\displaystyle \text{Let }E\text{ be the event that the two }T\text{'s come together.}
\displaystyle \text{Treat the two }T\text{'s as one block }(TT).
\displaystyle \text{Then there are }9\text{ distinct objects which can be arranged in }9!\text{ ways.}
\displaystyle \therefore n(E)=9!.
\displaystyle P(E)=\frac{n(E)}{n(S)}=\frac{9!}{\frac{10!}{2!}}=\frac{2\times9!}{10!}=\frac{1}{5}.
\displaystyle \therefore \text{The required probability is }\frac{1}{5}.
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{A committee of two persons is selected from two men and two women. Find the}
\displaystyle \text{probability that the committee has: (i) no man (ii) one man (iii) two men.}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of persons}=2+2=4.
\displaystyle \text{Number of ways of selecting }2\text{ persons from }4={}^{4}C_{2}.
\displaystyle \therefore n(S)={}^{4}C_{2}=6.
\displaystyle \text{(i) Let }E_1\text{ be the event that the committee has no man.}
\displaystyle n(E_1)={}^{2}C_{2}=1.
\displaystyle P(E_1)=\frac{{}^{2}C_{2}}{{}^{4}C_{2}}=\frac{1}{6}.
\displaystyle \text{(ii) Let }E_2\text{ be the event that the committee has one man.}
\displaystyle n(E_2)={}^{2}C_{1}\times{}^{2}C_{1}=2\times2=4.
\displaystyle P(E_2)=\frac{{}^{2}C_{1}\times{}^{2}C_{1}}{{}^{4}C_{2}}=\frac{4}{6}=\frac{2}{3}.
\displaystyle \text{(iii) Let }E_3\text{ be the event that the committee has two men.}
\displaystyle n(E_3)={}^{2}C_{2}=1.
\displaystyle P(E_3)=\frac{{}^{2}C_{2}}{{}^{4}C_{2}}=\frac{1}{6}.
\displaystyle \\

\displaystyle \textbf{Question 26: }\text{If the odds in favour of an event are }2:3,\text{ find the probability of the}
\displaystyle \text{occurrence of the event.}
\displaystyle \text{Answer:}
\displaystyle \text{Odds in favour of the event}=2:3.
\displaystyle \text{Let the favourable outcomes}=2k\text{ and the unfavourable outcomes}=3k.
\displaystyle \text{Total number of outcomes}=2k+3k=5k.
\displaystyle \therefore n(S)=5k.
\displaystyle \text{Let }E\text{ be the event that the event occurs.}
\displaystyle n(E)=2k.
\displaystyle P(E)=\frac{n(E)}{n(S)}=\frac{2k}{5k}=\frac{2}{5}.
\displaystyle \therefore \text{The required probability is }\frac{2}{5}.
\displaystyle \\

\displaystyle \textbf{Question 27: }\text{If the odds against an event are }7:9,\text{ find the probability of the}
\displaystyle \text{non-occurrence of the event.}
\displaystyle \text{Answer:}
\displaystyle \text{Odds against the event}=7:9.
\displaystyle \text{Let the unfavourable outcomes}=7k\text{ and the favourable outcomes}=9k.
\displaystyle \text{Total number of outcomes}=7k+9k=16k.
\displaystyle \therefore n(S)=16k.
\displaystyle \text{Let }E\text{ be the event that the event does not occur.}
\displaystyle n(E)=7k.
\displaystyle P(E)=\frac{n(E)}{n(S)}=\frac{7k}{16k}=\frac{7}{16}.
\displaystyle \therefore \text{The required probability of non-occurrence is }\frac{7}{16}.
\displaystyle \\

\displaystyle \textbf{Question 28: }\text{Two balls are drawn at random, one after another without replacement, from a bag}
\displaystyle \text{containing }2\text{ white, }3\text{ red, }5\text{ green and }4\text{ black balls. Find the probability that}
\displaystyle \text{the two balls are of different colours.}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of balls}=2+3+5+4=14.
\displaystyle \text{Total number of ways of selecting }2\text{ balls}={}^{14}C_2=91.
\displaystyle \therefore n(S)=91.
\displaystyle \text{Let }E\text{ be the event that the two selected balls are of different colours.}
\displaystyle n(E)={}^{2}C_1{}^{3}C_1+{}^{2}C_1{}^{5}C_1+{}^{2}C_1{}^{4}C_1
\displaystyle \qquad+{}^{3}C_1{}^{5}C_1+{}^{3}C_1{}^{4}C_1+{}^{5}C_1{}^{4}C_1.
\displaystyle n(E)=6+10+8+15+12+20=71.
\displaystyle P(E)=\frac{n(E)}{n(S)}=\frac{71}{91}.
\displaystyle \therefore \text{The required probability is }\frac{71}{91}.
\displaystyle \\

\displaystyle \textbf{Question 29: }\text{Two unbiased dice are thrown. Find the probability that:}
\displaystyle \text{(i) neither a doublet nor a total of }8\text{ appears}
\displaystyle \text{(ii) the sum of the numbers on the dice is neither a multiple of }2\text{ nor a multiple of }3.
\displaystyle \text{Answer:}
\displaystyle \text{Total number of possible outcomes}=6\times6=36.
\displaystyle \therefore n(S)=36.
\displaystyle \text{(i) Let }E\text{ be the event that neither a doublet nor a total of }8\text{ appears.}
\displaystyle \text{Then }E'\text{ is the event that a doublet or a total of }8\text{ appears.}
\displaystyle \text{Number of doublets}=6.
\displaystyle \text{The outcomes having a total of }8\text{ are}
\displaystyle (2,6),(3,5),(4,4),(5,3),(6,2).
\displaystyle \text{Thus, the number of outcomes having a total of }8=5.
\displaystyle \text{The outcome }(4,4)\text{ is common to both events.}
\displaystyle \therefore n(E')=6+5-1=10.
\displaystyle P(E')=\frac{10}{36}=\frac{5}{18}.
\displaystyle P(E)=1-P(E')=1-\frac{5}{18}=\frac{13}{18}.
\displaystyle \therefore \text{The required probability is }\frac{13}{18}.
\displaystyle \text{(ii) Let }F\text{ be the event that the sum is neither a multiple of }2\text{ nor a multiple of }3.
\displaystyle \text{The possible sums satisfying the condition are }5,\ 7\text{ and }11.
\displaystyle \text{Number of outcomes having a sum of }5=4.
\displaystyle \text{Number of outcomes having a sum of }7=6.
\displaystyle \text{Number of outcomes having a sum of }11=2.
\displaystyle \therefore n(F)=4+6+2=12.
\displaystyle P(F)=\frac{n(F)}{n(S)}=\frac{12}{36}=\frac{1}{3}.
\displaystyle \therefore \text{The required probability is }\frac{1}{3}.
\displaystyle \\

\displaystyle \textbf{Question 30: }\text{A bag contains }8\text{ red, }3\text{ white and }9\text{ blue balls. If three balls are}
\displaystyle \text{drawn at random, determine the probability that:}
\displaystyle \text{(i) all three balls are blue}\qquad\text{(ii) all three balls are of different colours.}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of balls}=8+3+9=20.
\displaystyle \text{Total number of ways of selecting }3\text{ balls}={}^{20}C_3=1140.
\displaystyle \therefore n(S)=1140.
\displaystyle \text{(i) Let }E_1\text{ be the event that all three balls are blue.}
\displaystyle n(E_1)={}^{9}C_3=84.
\displaystyle P(E_1)=\frac{n(E_1)}{n(S)}=\frac{84}{1140}=\frac{7}{95}.
\displaystyle \text{(ii) Let }E_2\text{ be the event that all three balls are of different colours.}
\displaystyle n(E_2)={}^{8}C_1\times{}^{3}C_1\times{}^{9}C_1=8\times3\times9=216.
\displaystyle P(E_2)=\frac{n(E_2)}{n(S)}=\frac{216}{1140}=\frac{18}{95}.
\displaystyle \therefore \text{The required probabilities are }\frac{7}{95}\text{ and }\frac{18}{95}\text{ respectively.}
\displaystyle \\

\displaystyle \textbf{Question 31: }\text{A bag contains }5\text{ red, }5\text{ white and }7\text{ black balls. Two balls are}
\displaystyle \text{drawn at random. Find the probability that both balls are red or both are black.}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of balls}=5+5+7=17.
\displaystyle \text{Total number of ways of selecting }2\text{ balls}={}^{17}C_2=136.
\displaystyle \therefore n(S)=136.
\displaystyle \text{Let }E\text{ be the event that both balls are red or both are black.}
\displaystyle n(E)={}^{5}C_2+{}^{7}C_2=10+21=31.
\displaystyle P(E)=\frac{n(E)}{n(S)}=\frac{31}{136}.
\displaystyle \therefore \text{The required probability is }\frac{31}{136}.
\displaystyle \\

\displaystyle \textbf{Question 32: }\text{If a letter is chosen at random from the English alphabet, find the probability}
\displaystyle \text{that the letter is (i) a vowel (ii) a consonant.}
\displaystyle \text{Answer:}
\displaystyle \text{The English alphabet contains }26\text{ letters.}
\displaystyle \therefore n(S)={}^{26}C_1=26.
\displaystyle \text{(i) Let }E_1\text{ be the event that the chosen letter is a vowel.}
\displaystyle \text{The vowels are }a,\ e,\ i,\ o,\ u.
\displaystyle n(E_1)=5.
\displaystyle P(E_1)=\frac{n(E_1)}{n(S)}=\frac{5}{26}.
\displaystyle \text{(ii) Let }E_2\text{ be the event that the chosen letter is a consonant.}
\displaystyle n(E_2)=26-5=21.
\displaystyle P(E_2)=\frac{n(E_2)}{n(S)}=\frac{21}{26}.
\displaystyle \therefore \text{The required probabilities are }\frac{5}{26}\text{ and }\frac{21}{26}\text{ respectively.}
\displaystyle \\

\displaystyle \textbf{Question 33: }\text{In a lottery, a person chooses six different numbers from }1\text{ to }20\text{ at random.}
\displaystyle \text{If these match the six numbers fixed by the lottery committee, he wins the prize.}
\displaystyle \text{Find the probability of winning the prize.}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of ways of choosing }6\text{ numbers from }20={}^{20}C_6=38760.
\displaystyle \therefore n(S)=38760.
\displaystyle \text{Let }E\text{ be the event that all the chosen numbers match the fixed winning numbers.}
\displaystyle \text{Since only one combination is the winning combination, }n(E)=1.
\displaystyle P(E)=\frac{n(E)}{n(S)}=\frac{1}{{}^{20}C_6}=\frac{1}{38760}.
\displaystyle \therefore \text{The probability of winning the prize is }\frac{1}{38760}.
\displaystyle \\

\displaystyle \textbf{Question 34: }\text{Twenty cards are numbered from }1\text{ to }20.\text{ One card is drawn at random.}
\displaystyle \text{Find the probability that the number on the card is:}
\displaystyle \text{(i) a multiple of }4\qquad\text{(ii) not a multiple of }4\qquad\text{(iii) odd}
\displaystyle \text{(iv) greater than }12\qquad\text{(v) divisible by }5\qquad\text{(vi) not a multiple of }6.
\displaystyle \text{Answer:}
\displaystyle \text{Total number of cards}=20.
\displaystyle \therefore n(S)={}^{20}C_1=20.
\displaystyle \text{(i) Let }E_1\text{ be the event that the number is a multiple of }4.
\displaystyle n(E_1)=5\text{ i.e. }\{4,8,12,16,20\}.
\displaystyle P(E_1)=\frac{n(E_1)}{n(S)}=\frac{5}{20}=\frac{1}{4}.
\displaystyle \text{(ii) Let }E_2\text{ be the event that the number is not a multiple of }4.
\displaystyle \text{Then }E_2'\text{ is the event that the number is a multiple of }4.
\displaystyle n(E_2')=5.
\displaystyle P(E_2')=\frac{n(E_2')}{n(S)}=\frac{5}{20}=\frac{1}{4}.
\displaystyle P(E_2)=1-P(E_2')=1-\frac{1}{4}=\frac{3}{4}.
\displaystyle \text{(iii) Let }E_3\text{ be the event that the number is odd.}
\displaystyle n(E_3)=10\text{ i.e. }\{1,3,5,7,9,11,13,15,17,19\}.
\displaystyle P(E_3)=\frac{n(E_3)}{n(S)}=\frac{10}{20}=\frac{1}{2}.
\displaystyle \text{(iv) Let }E_4\text{ be the event that the number is greater than }12.
\displaystyle n(E_4)=8\text{ i.e. }\{13,14,15,16,17,18,19,20\}.
\displaystyle P(E_4)=\frac{n(E_4)}{n(S)}=\frac{8}{20}=\frac{2}{5}.
\displaystyle \text{(v) Let }E_5\text{ be the event that the number is divisible by }5.
\displaystyle n(E_5)=4\text{ i.e. }\{5,10,15,20\}.
\displaystyle P(E_5)=\frac{n(E_5)}{n(S)}=\frac{4}{20}=\frac{1}{5}.
\displaystyle \text{(vi) Let }E_6\text{ be the event that the number is not a multiple of }6.
\displaystyle \text{Then }E_6'\text{ is the event that the number is a multiple of }6.
\displaystyle n(E_6')=3\text{ i.e. }\{6,12,18\}.
\displaystyle P(E_6')=\frac{n(E_6')}{n(S)}=\frac{3}{20}.
\displaystyle P(E_6)=1-P(E_6')=1-\frac{3}{20}=\frac{17}{20}.
\displaystyle \\

\displaystyle \textbf{Question 35: }\text{Two dice are thrown. Find the odds in favour of getting the sum}
\displaystyle \text{(i) }4\qquad\text{(ii) }5\qquad\text{(iii) \text{Find the odds against getting the sum }6.}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of possible outcomes}=6\times6=36.
\displaystyle \text{(i) Let }E_1\text{ be the event that the sum is }4.
\displaystyle \text{Favourable outcomes}=\{(1,3),(2,2),(3,1)\}.
\displaystyle n(E_1)=3.
\displaystyle \text{Number of unfavourable outcomes}=36-3=33.
\displaystyle \therefore \text{Odds in favour of getting the sum }4=3:33=1:11.
\displaystyle \text{(ii) Let }E_2\text{ be the event that the sum is }5.
\displaystyle \text{Favourable outcomes}=\{(1,4),(2,3),(3,2),(4,1)\}.
\displaystyle n(E_2)=4.
\displaystyle \text{Number of unfavourable outcomes}=36-4=32.
\displaystyle \therefore \text{Odds in favour of getting the sum }5=4:32=1:8.
\displaystyle \text{(iii) Let }E_3\text{ be the event that the sum is }6.
\displaystyle \text{Favourable outcomes}=\{(1,5),(2,4),(3,3),(4,2),(5,1)\}.
\displaystyle n(E_3)=5.
\displaystyle \text{Number of unfavourable outcomes}=36-5=31.
\displaystyle \therefore \text{Odds against getting the sum }6=31:5.
\displaystyle \\

\displaystyle \textbf{Question 36: }\text{What are the odds in favour of getting a spade if a card is drawn from a}
\displaystyle \text{well-shuffled deck of }52\text{ cards? What are the odds in favour of getting a king?}
\displaystyle \text{Answer:}
\displaystyle \text{A standard deck contains }52\text{ cards.}
\displaystyle \text{(i) Let }E_1\text{ be the event of drawing a spade.}
\displaystyle \text{Number of favourable outcomes}=13.
\displaystyle \text{Number of unfavourable outcomes}=52-13=39.
\displaystyle \therefore \text{Odds in favour of getting a spade}=13:39=1:3.
\displaystyle \text{(ii) Let }E_2\text{ be the event of drawing a king.}
\displaystyle \text{Number of favourable outcomes}=4.
\displaystyle \text{Number of unfavourable outcomes}=52-4=48.
\displaystyle \therefore \text{Odds in favour of getting a king}=4:48=1:12.
\displaystyle \\

\displaystyle \textbf{Question 37: }\text{A box contains }10\text{ red, }20\text{ blue and }30\text{ green marbles. Five marbles}
\displaystyle \text{are drawn at random. Find the probability that:}
\displaystyle \text{(i) all are blue}\qquad\text{(ii) at least one is green.}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of marbles}=10+20+30=60.
\displaystyle \text{Total number of ways of selecting }5\text{ marbles}={}^{60}C_5.
\displaystyle \therefore n(S)={}^{60}C_5.
\displaystyle \text{(i) Let }E_1\text{ be the event that all five selected marbles are blue.}
\displaystyle n(E_1)={}^{20}C_5.
\displaystyle P(E_1)=\frac{n(E_1)}{n(S)}=\frac{{}^{20}C_5}{{}^{60}C_5}.
\displaystyle =\frac{20\times19\times18\times17\times16}{60\times59\times58\times57\times56}.
\displaystyle =\frac{34}{11977}.
\displaystyle \therefore \text{The probability that all five marbles are blue is }\frac{34}{11977}.
\displaystyle \text{(ii) Let }E_2\text{ be the event that at least one selected marble is green.}
\displaystyle \text{Then }E_2'\text{ is the event that no selected marble is green.}
\displaystyle \text{The number of non-green marbles}=10+20=30.
\displaystyle n(E_2')={}^{30}C_5.
\displaystyle P(E_2')=\frac{n(E_2')}{n(S)}=\frac{{}^{30}C_5}{{}^{60}C_5}.
\displaystyle P(E_2)=1-P(E_2')=1-\frac{{}^{30}C_5}{{}^{60}C_5}.
\displaystyle =1-\frac{30\times29\times28\times27\times26}{60\times59\times58\times57\times56}.
\displaystyle =1-\frac{117}{4484}=\frac{4367}{4484}.
\displaystyle \therefore \text{The probability that at least one marble is green is }\frac{4367}{4484}.
\displaystyle \\

\displaystyle \textbf{Question 38: }\text{A box contains }6\text{ red marbles numbered }1\text{ to }6\text{ and }4\text{ white marbles}
\displaystyle \text{numbered }12\text{ to }15.\text{ One marble is drawn at random. Find the probability that it is:}
\displaystyle \text{(i) white}\qquad\text{(ii) white and odd-numbered}\qquad\text{(iii) even-numbered}
\displaystyle \text{(iv) red or even-numbered.}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of marbles}=6+4=10.
\displaystyle \therefore n(S)={}^{10}C_1=10.
\displaystyle \text{(i) Let }E_1\text{ be the event of drawing a white marble.}
\displaystyle n(E_1)=4.
\displaystyle P(E_1)=\frac{n(E_1)}{n(S)}=\frac{4}{10}=\frac{2}{5}.
\displaystyle \text{(ii) Let }E_2\text{ be the event of drawing a white odd-numbered marble.}
\displaystyle n(E_2)=2\text{ i.e. }\{13,15\}.
\displaystyle P(E_2)=\frac{n(E_2)}{n(S)}=\frac{2}{10}=\frac{1}{5}.
\displaystyle \text{(iii) Let }E_3\text{ be the event of drawing an even-numbered marble.}
\displaystyle n(E_3)=5\text{ i.e. }\{2,4,6,12,14\}.
\displaystyle P(E_3)=\frac{n(E_3)}{n(S)}=\frac{5}{10}=\frac{1}{2}.
\displaystyle \text{(iv) Let }E_4\text{ be the event of drawing a red marble or an even-numbered marble.}
\displaystyle n(E_4)=8\text{ i.e. }\{1,2,3,4,5,6,12,14\}.
\displaystyle P(E_4)=\frac{n(E_4)}{n(S)}=\frac{8}{10}=\frac{4}{5}.
\displaystyle \therefore \text{The required probabilities are }\frac{2}{5},\ \frac{1}{5},\ \frac{1}{2}\text{ and }\frac{4}{5}\text{ respectively.}
\displaystyle \\

\displaystyle \textbf{Question 39: }\text{A class consists of }10\text{ boys and }8\text{ girls. Three students are selected at}
\displaystyle \text{random. Find the probability that the selected group has:}
\displaystyle \text{(i) all boys}\qquad\text{(ii) all girls}\qquad\text{(iii) one boy and two girls}
\displaystyle \text{(iv) at least one girl}\qquad\text{(v) at most one girl.}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of students}=10+8=18.
\displaystyle \text{Total number of ways of selecting }3\text{ students}={}^{18}C_3=816.
\displaystyle \therefore n(S)=816.
\displaystyle \text{(i) Let }E_1\text{ be the event that all three selected students are boys.}
\displaystyle n(E_1)={}^{10}C_3=120.
\displaystyle P(E_1)=\frac{n(E_1)}{n(S)}=\frac{120}{816}=\frac{5}{34}.
\displaystyle \text{(ii) Let }E_2\text{ be the event that all three selected students are girls.}
\displaystyle n(E_2)={}^{8}C_3=56.
\displaystyle P(E_2)=\frac{n(E_2)}{n(S)}=\frac{56}{816}=\frac{7}{102}.
\displaystyle \text{(iii) Let }E_3\text{ be the event that one boy and two girls are selected.}
\displaystyle n(E_3)={}^{10}C_1\times{}^{8}C_2=10\times28=280.
\displaystyle P(E_3)=\frac{n(E_3)}{n(S)}=\frac{280}{816}=\frac{35}{102}.
\displaystyle \text{(iv) Let }E_4\text{ be the event that at least one girl is selected.}
\displaystyle \text{Then }E_4'\text{ is the event that all three selected students are boys.}
\displaystyle P(E_4')=\frac{{}^{10}C_3}{{}^{18}C_3}=\frac{120}{816}=\frac{5}{34}.
\displaystyle P(E_4)=1-P(E_4')=1-\frac{5}{34}=\frac{29}{34}.
\displaystyle \text{(v) Let }E_5\text{ be the event that at most one girl is selected.}
\displaystyle \text{Thus, either no girl or exactly one girl is selected.}
\displaystyle n(E_5)={}^{10}C_3+{}^{10}C_2\times{}^{8}C_1.
\displaystyle =120+45\times8=120+360=480.
\displaystyle P(E_5)=\frac{n(E_5)}{n(S)}=\frac{480}{816}=\frac{10}{17}.
\displaystyle \therefore \text{The required probabilities are }\frac{5}{34},\ \frac{7}{102},\ \frac{35}{102},\ \frac{29}{34}\text{ and }\frac{10}{17}.
\displaystyle \\

\displaystyle \textbf{Question 40: }\text{Five cards are drawn from a well-shuffled pack of }52\text{ cards. Find the}
\displaystyle \text{probability that all the five cards are hearts.}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of ways of drawing }5\text{ cards from }52={}^{52}C_5=2598960.
\displaystyle \therefore n(S)=2598960.
\displaystyle \text{Let }E\text{ be the event that all the five cards are hearts.}
\displaystyle n(E)={}^{13}C_5=1287.
\displaystyle P(E)=\frac{n(E)}{n(S)}=\frac{{}^{13}C_5}{{}^{52}C_5}=\frac{1287}{2598960}=\frac{33}{66640}.
\displaystyle \therefore \text{The required probability is }\frac{33}{66640}.
\displaystyle \\

\displaystyle \textbf{Question 41: }\text{A bag contains tickets numbered from }1\text{ to }20.\text{ Two tickets are drawn at}
\displaystyle \text{random. Find the probability that:}
\displaystyle \text{(i) both tickets have prime numbers}\qquad\text{(ii) one ticket has a prime number and}
\displaystyle \text{the other has a multiple of }4.
\displaystyle \text{Answer:}
\displaystyle \text{Total number of ways of drawing }2\text{ tickets from }20={}^{20}C_2=190.
\displaystyle \therefore n(S)=190.
\displaystyle \text{(i) The prime numbers from }1\text{ to }20\text{ are }\{2,3,5,7,11,13,17,19\}.
\displaystyle \text{Hence there are }8\text{ prime numbers.}
\displaystyle n(E_1)={}^{8}C_2=28.
\displaystyle P(E_1)=\frac{n(E_1)}{n(S)}=\frac{28}{190}=\frac{14}{95}.
\displaystyle \text{(ii) The multiples of }4\text{ are }\{4,8,12,16,20\}.
\displaystyle \text{Hence there are }5\text{ multiples of }4.
\displaystyle n(E_2)={}^{8}C_1\times{}^{5}C_1=8\times5=40.
\displaystyle P(E_2)=\frac{n(E_2)}{n(S)}=\frac{40}{190}=\frac{4}{19}.
\displaystyle \therefore \text{The required probabilities are }\frac{14}{95}\text{ and }\frac{4}{19}\text{ respectively.}
\displaystyle \\

\displaystyle \textbf{Question 42: }\text{An urn contains }7\text{ white, }5\text{ black and }3\text{ red balls. Two balls are}
\displaystyle \text{drawn at random. Find the probability that:}
\displaystyle \text{(i) both the balls are red}\qquad\text{(ii) one ball is red and the other is black}
\displaystyle \text{(iii) exactly one ball is white.}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of balls}=7+5+3=15.
\displaystyle \text{Total number of ways of drawing }2\text{ balls}={}^{15}C_2=105.
\displaystyle \therefore n(S)=105.
\displaystyle \text{(i) Let }E_1\text{ be the event that both the balls are red.}
\displaystyle n(E_1)={}^{3}C_2=3.
\displaystyle P(E_1)=\frac{n(E_1)}{n(S)}=\frac{3}{105}=\frac{1}{35}.
\displaystyle \text{(ii) Let }E_2\text{ be the event that one ball is red and the other is black.}
\displaystyle n(E_2)={}^{3}C_1\times{}^{5}C_1=3\times5=15.
\displaystyle P(E_2)=\frac{n(E_2)}{n(S)}=\frac{15}{105}=\frac{1}{7}.
\displaystyle \text{(iii) Let }E_3\text{ be the event that exactly one ball is white.}
\displaystyle \text{Choose }1\text{ white ball from }7\text{ and }1\text{ non-white ball from }8.
\displaystyle n(E_3)={}^{7}C_1\times{}^{8}C_1=7\times8=56.
\displaystyle P(E_3)=\frac{n(E_3)}{n(S)}=\frac{56}{105}=\frac{8}{15}.
\displaystyle \therefore \text{The required probabilities are }\frac{1}{35},\ \frac{1}{7}\text{ and }\frac{8}{15}\text{ respectively.}
\displaystyle \\

\displaystyle \textbf{Question 43: }\text{A and B each throw a pair of dice. If A throws a total of }9,\text{ find the}
\displaystyle \text{probability that B throws a higher total.}
\displaystyle \text{Answer:}
\displaystyle \text{When B throws a pair of dice, the total number of possible outcomes}=6\times6=36.
\displaystyle \therefore n(S)=36.
\displaystyle \text{Let }E\text{ be the event that B throws a total greater than }9.
\displaystyle \text{The favourable outcomes are}
\displaystyle \{(4,6),(5,5),(6,4),(5,6),(6,5),(6,6)\}.
\displaystyle \therefore n(E)=6.
\displaystyle P(E)=\frac{n(E)}{n(S)}=\frac{6}{36}=\frac{1}{6}.
\displaystyle \therefore \text{B's chance of throwing a higher total is }\frac{1}{6}.
\displaystyle \\

\displaystyle \textbf{Question 44: }\text{In a hand at Whist, find the probability that all four kings are held by a}
\displaystyle \text{specified player.}
\displaystyle \text{Answer:}
\displaystyle \text{A specified player receives }13\text{ cards in a game of Whist.}
\displaystyle \text{Total number of possible }13\text{-card hands}={}^{52}C_{13}.
\displaystyle \therefore n(S)={}^{52}C_{13}.
\displaystyle \text{Let }E\text{ be the event that the specified player holds all four kings.}
\displaystyle \text{After choosing the }4\text{ kings, the remaining }9\text{ cards are chosen from }48\text{ cards.}
\displaystyle \therefore n(E)={}^{48}C_9.
\displaystyle P(E)=\frac{n(E)}{n(S)}=\frac{{}^{48}C_9}{{}^{52}C_{13}}.
\displaystyle =\frac{13\times12\times11\times10}{52\times51\times50\times49}=\frac{11}{4165}.
\displaystyle \therefore \text{The required probability is }\frac{11}{4165}.
\displaystyle \\

\displaystyle \textbf{Question 45: }\text{Find the probability that in a random arrangement of the letters of the word}
\displaystyle \text{`UNIVERSITY', the two }I\text{'s do not come together.}
\displaystyle \text{Answer:}
\displaystyle \text{The word UNIVERSITY contains }10\text{ letters, with }I\text{ repeated twice.}
\displaystyle \text{Total number of distinct arrangements}=\frac{10!}{2!}.
\displaystyle \therefore n(S)=\frac{10!}{2!}.
\displaystyle \text{Let }E\text{ be the event that the two }I\text{'s come together.}
\displaystyle \text{Treat the two }I\text{'s as one block.}
\displaystyle \text{Then there are }9\text{ distinct objects which can be arranged in }9!\text{ ways.}
\displaystyle \therefore n(E)=9!.
\displaystyle \text{Number of arrangements in which the two }I\text{'s do not come together}=\frac{10!}{2!}-9!.
\displaystyle \therefore P(\text{two }I\text{'s do not come together})=\frac{\frac{10!}{2!}-9!}{\frac{10!}{2!}}.
\displaystyle =1-\frac{2\times9!}{10!}=1-\frac{1}{5}=\frac{4}{5}.
\displaystyle \therefore \text{The required probability is }\frac{4}{5}.
\displaystyle \\

 

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