\displaystyle \textbf{Question 1. }\text{Three numbers are chosen at random from numbers }1\text{ to }30.\text{ Write the} \\ \text{probability that the chosen numbers are consecutive.}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of ways of choosing 3 numbers from 30 numbers}
\displaystyle ={}^{30}C_3
\displaystyle =\frac{30\times29\times28}{3\times2\times1}=4060
\displaystyle \text{Consecutive sets are }(1,2,3),(2,3,4),\ldots,(28,29,30)
\displaystyle \therefore \text{Number of favourable cases}=28
\displaystyle \therefore P(\text{consecutive numbers})=\frac{28}{{}^{30}C_3}
\displaystyle =\frac{28}{4060}
\displaystyle =\frac{1}{145}
\\

\displaystyle \textbf{Question 2. }n(>3)\text{ persons are sitting in a row. Two of them are selected. Write the} \\ \text{probability that they are together.}
\displaystyle \text{Answer:}
\displaystyle \text{Total ways of selecting }2\text{ persons from }n\text{ persons}={}^{n}C_2
\displaystyle \text{Number of adjacent pairs}=(n-1)
\displaystyle \therefore P=\frac{n-1}{{}^{n}C_2}=\frac{n-1}{\frac{n(n-1)}{2}}=\frac{2}{n}
\\

\displaystyle \textbf{Question 3. }\text{A single letter is selected at random from the word PROBABILITY. What is} \\ \text{the probability that it is a vowel?}
\displaystyle \text{Answer:}
\displaystyle \text{The word PROBABILITY has }11\text{ letters.}
\displaystyle \text{Vowels are }O,A,I,I
\displaystyle \text{Number of vowels}=4
\displaystyle \therefore P=\frac{4}{11}
\\

\displaystyle \textbf{Question 4. }\text{What is the probability that a leap year will have }53\text{ Fridays or }53\text{ Saturdays?}
\displaystyle \text{Answer:}
\displaystyle \text{A leap year has }366\text{ days}=52\text{ weeks}+2\text{ days.}
\displaystyle \text{The extra two days can be }
\displaystyle (Sun,Mon),(Mon,Tue),(Tue,Wed),(Wed,Thu),(Thu,Fri),(Fri,Sat),(Sat,Sun)
\displaystyle \text{For }53\text{ Fridays or }53\text{ Saturdays, favourable cases are}
\displaystyle (Thu,Fri),(Fri,Sat),(Sat,Sun)
\displaystyle \therefore P=\frac{3}{7}
\\

\displaystyle \textbf{Question 5. }\text{Three dice are thrown simultaneously. What is the probability of getting }15\text{ as} \\ \text{the sum?}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of outcomes}=6^3=216
\displaystyle \text{Possible triples whose sum is }15\text{ are}
\displaystyle (6,6,3),(6,3,6),(3,6,6)
\displaystyle (6,5,4),(6,4,5),(5,6,4),(5,4,6),(4,6,5),(4,5,6)
\displaystyle (5,5,5)
\displaystyle \therefore \text{Number of favourable outcomes}=10
\displaystyle \therefore P(\text{sum }15)=\frac{10}{216}=\frac{5}{108}
\\

\displaystyle \textbf{Question 6. }\text{If the letters of the word 'MISSISSIPPI' are written down at random in} \\ \text{a row, what is the probability that four S's come together.}
\displaystyle \text{Answer:}
\displaystyle \text{The word MISSISSIPPI has }11\text{ letters with }4I,4S,2P,1M
\displaystyle \text{Total number of arrangements}=\frac{11!}{4!4!2!}
\displaystyle \text{If all four S's come together, treat SSSS as one object}
\displaystyle \text{Then we have }8\text{ objects: }(SSSS),4I,2P,1M
\displaystyle \text{Number of favourable arrangements}=\frac{8!}{4!2!}
\displaystyle \therefore P(\text{four S's together})=\frac{\frac{8!}{4!2!}}{\frac{11!}{4!4!2!}}
\displaystyle =\frac{8!\times4!}{11!}
\displaystyle =\frac{24}{11\times10\times9}
\displaystyle =\frac{4}{165}
\\

\displaystyle \textbf{Question 7. }\text{What is the probability that the 13th day of a randomly chosen} \\ \text{month is Friday?}
\displaystyle \text{Answer:}
\displaystyle \text{Any date of a month is equally likely to fall on any one of the 7 days of the week.}
\displaystyle \text{The 13th day can therefore be Monday, Tuesday, }\ldots\text{ or Sunday.}
\displaystyle \text{Number of favourable outcomes}=1
\displaystyle \text{Total number of possible outcomes}=7
\displaystyle \therefore P(\text{13th day is Friday})=\frac{1}{7}
\\

\displaystyle \textbf{Question 8. }\text{Three of the six vertices of a regular hexagon are chosen at random.} \\ \text{What is the probability that the triangle with these vertices is equilateral?}
\displaystyle \text{Answer:}
\displaystyle \text{Total ways of choosing }3\text{ vertices}={}^{6}C_3=20
\displaystyle \text{Equilateral triangles are formed by alternate vertices.}
\displaystyle \text{Favourable triangles}=2
\displaystyle \therefore P=\frac{2}{20}=\frac{1}{10}
\\

\displaystyle \textbf{Question 9. }\text{If }E_1\text{ and }E_2\text{ are independent events, write the value of } \\ P\left((E_1\cup E_2)\cap(\overline{E_1}\cap \overline{E_2})\right).
\displaystyle \text{Answer:}
\displaystyle (E_1\cup E_2)\cap(\overline{E_1}\cap \overline{E_2})=\phi
\displaystyle \therefore P\left((E_1\cup E_2)\cap(\overline{E_1}\cap \overline{E_2})\right)=0
\\

\displaystyle \textbf{Question 10. }\text{If }A\text{ and }B\text{ are two independent events such that } \\ P(A\cap B)=\frac{1}{6}\text{ and }P(\overline{A}\cap \overline{B})=\frac{1}{3},\text{ then write the values of }P(A)\text{ and }P(B).
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(A)=x\text{ and }P(B)=y
\displaystyle \text{Since }A\text{ and }B\text{ are independent,}
\displaystyle xy=P(A\cap B)=\frac{1}{6}
\displaystyle P(\overline{A}\cap \overline{B})=P(\overline{A})P(\overline{B})
\displaystyle (1-x)(1-y)=\frac{1}{3}
\displaystyle 1-x-y+xy=\frac{1}{3}
\displaystyle 1-x-y+\frac{1}{6}=\frac{1}{3}
\displaystyle x+y=\frac{5}{6}
\displaystyle \text{Now, }x\text{ and }y\text{ are roots of }t^2-\frac{5}{6}t+\frac{1}{6}=0
\displaystyle 6t^2-5t+1=0
\displaystyle (3t-1)(2t-1)=0
\displaystyle t=\frac{1}{3}\text{ or }\frac{1}{2}
\displaystyle \therefore P(A)=\frac{1}{2},\quad P(B)=\frac{1}{3}\text{ or vice versa.}
\\

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