\displaystyle \textbf{Question 1: }\text{An experiment in which all the outcomes are known in advance and the exact}
\displaystyle \text{outcome is not known in advance is called a }\underline{\hspace{2cm}}\text{ experiment.}
\displaystyle \text{Answer:}
\displaystyle \text{Random experiment.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Outcomes of a random experiment are called }\underline{\hspace{2cm}}\text{ events.}
\displaystyle \text{Answer:}
\displaystyle \text{Elementary events.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{The set of all possible outcomes of a random experiment is called the}
\displaystyle \underline{\hspace{2cm}}\text{.}
\displaystyle \text{Answer:}
\displaystyle \text{Sample space.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Subsets of a sample space are called }\underline{\hspace{2cm}}\text{.}
\displaystyle \text{Answer:}
\displaystyle \text{Events.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{An event obtained by combining two or more elementary events is called a}
\displaystyle \underline{\hspace{2cm}}\text{ event.}
\displaystyle \text{Answer:}
\displaystyle \text{Compound event.}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{The range of probability of any event of a random experiment is}
\displaystyle \underline{\hspace{2cm}}\text{.}
\displaystyle \text{Answer:}
\displaystyle 0\leq P(A)\leq1.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{The probability of any event associated to a random experiment cannot}
\displaystyle \text{exceed }\underline{\hspace{2cm}}\text{.}
\displaystyle \text{Answer:}
\displaystyle 1.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{If }A^c\text{ or }\bar{A}\text{ or }A'\text{ denote the complement of an event }A,\text{ then}
\displaystyle P(A^c)=\underline{\hspace{2cm}}\text{.}
\displaystyle \text{Answer:}
\displaystyle P(A)+P(A^c)=1.
\displaystyle \therefore P(A^c)=1-P(A).
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{If an event }A\text{ cannot occur, then }P(A)=\underline{\hspace{2cm}}\text{.}
\displaystyle \text{Answer:}
\displaystyle \text{If an event cannot occur, it is an impossible event.}
\displaystyle \therefore P(A)=0.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{The probability expressed as a percentage of a particular occurrence can}
\displaystyle \text{never be less than }\underline{\hspace{1.5cm}}\text{ and greater than }\underline{\hspace{1.5cm}}\text{.}
\displaystyle \text{Answer:}
\displaystyle 0\leq P(A)\leq1.
\displaystyle \therefore 0\%\leq P(A)\leq100\%.
\displaystyle \therefore \text{The required entries are }0\%\text{ and }100\%\text{.}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Getting any one of the numbers from }1\text{ to }6\text{ on the upper face of a die when}
\displaystyle \text{thrown is called }\underline{\hspace{2cm}}\text{ event.}
\displaystyle \text{Answer:}
\displaystyle \text{Elementary event.}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{Probability of a sure event is }\underline{\hspace{2cm}}\text{.}
\displaystyle \text{Answer:}
\displaystyle P(\text{sure event})=1.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{Probability of an impossible event is }\underline{\hspace{2cm}}\text{.}
\displaystyle \text{Answer:}
\displaystyle P(\text{impossible event})=0.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{The probability of an event (other than sure and impossible event) lies}
\displaystyle \text{between }\underline{\hspace{2cm}}\text{.}
\displaystyle \text{Answer:}
\displaystyle 0<P(A)<1.
\displaystyle \therefore \text{The required entries are }0\text{ and }1.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{Every elementary event associated to a random experiment has}
\displaystyle \underline{\hspace{2cm}}\text{ probability.}
\displaystyle \text{Answer:}
\displaystyle \text{Non-negative probability and equal probability.}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{Probability of an event }A+\text{ Probability of event `not }A\text{'}
\displaystyle =\underline{\hspace{2cm}}\text{.}
\displaystyle \text{Answer:}
\displaystyle P(A)+P(\bar{A})=1.
\displaystyle \therefore \text{The required answer is }1.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{Sum of the probabilities of each outcome in an experiment is}
\displaystyle \underline{\hspace{2cm}}\text{.}
\displaystyle \text{Answer:}
\displaystyle \text{Sum of the probabilities of all possible outcomes}=1.
\displaystyle \therefore \text{The required answer is }1.
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{An urn contains }9\text{ red balls and }x\text{ green balls. If the probability of}
\displaystyle \text{picking a red ball is thrice that of picking a green ball, then }x=\underline{\hspace{2cm}}\text{.}
\displaystyle \text{Answer:}
\displaystyle P(\text{red})=\frac{9}{9+x},\qquad P(\text{green})=\frac{x}{9+x}.
\displaystyle \frac{9}{9+x}=3\left(\frac{x}{9+x}\right).
\displaystyle 9=3x.
\displaystyle \therefore x=3.
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{An urn contains }6\text{ blue and }x\text{ green balls. If the probability of drawing a}
\displaystyle \text{green ball is double that of drawing a blue ball, then }x=\underline{\hspace{2cm}}\text{.}
\displaystyle \text{Answer:}
\displaystyle P(\text{green})=\frac{x}{6+x},\qquad P(\text{blue})=\frac{6}{6+x}.
\displaystyle \frac{x}{6+x}=2\left(\frac{6}{6+x}\right).
\displaystyle x=12.
\displaystyle \therefore \text{The required answer is }12.
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{If one natural number is selected from the first }70\text{ natural numbers, the}
\displaystyle \text{probability that it is a solution of the inequation }3x-2>x+42,\text{ is }\underline{\hspace{2cm}}\text{.}
\displaystyle \text{Answer:}
\displaystyle 3x-2>x+42.
\displaystyle 2x>44.
\displaystyle x>22.
\displaystyle \text{Thus, favourable natural numbers are }23,24,\ldots,70.
\displaystyle \text{Number of favourable outcomes}=70-23+1=48.
\displaystyle \therefore P(x>22)=\frac{48}{70}=\frac{24}{35}.
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{Two dice are thrown together, the probability of getting a doublet is}
\displaystyle \underline{\hspace{2cm}}\text{.}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of elementary events}=6\times6=36.
\displaystyle \text{Doublets are }(1,1),(2,2),(3,3),(4,4),(5,5),(6,6).
\displaystyle \therefore P(\text{a doublet})=\frac{6}{36}=\frac{1}{6}.
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{Three dice are thrown together, the probability of getting the same number on}
\displaystyle \text{all the dice is }\underline{\hspace{2cm}}\text{.}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of elementary events}=6^3=216.
\displaystyle \text{Favourable outcomes are }(1,1,1),(2,2,2),\ldots,(6,6,6).
\displaystyle \therefore \text{Number of favourable outcomes}=6.
\displaystyle \therefore P(\text{same number on all three dice})=\frac{6}{216}=\frac{1}{36}.
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{A number is selected from the set }\{1,2,3,4,5,6,7,8\}.\text{ The probability that}
\displaystyle \text{it is a root of the equation }x^2-6x+8=0,\text{ is }\underline{\hspace{2cm}}\text{.}
\displaystyle \text{Answer:}
\displaystyle x^2-6x+8=0.
\displaystyle (x-2)(x-4)=0.
\displaystyle \therefore x=2\text{ or }4.
\displaystyle \text{Number of favourable outcomes}=2,\qquad\text{Total number of outcomes}=8.
\displaystyle \therefore P(\text{a root of the equation})=\frac{2}{8}=\frac{1}{4}.
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{A girl calculates that the probability of her winning the first prize in a lottery}
\displaystyle \text{is }0.08.\text{ If }6000\text{ tickets are sold, then the number of tickets she bought is }\underline{\hspace{2cm}}\text{.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the number of tickets bought by her}=x.
\displaystyle \frac{x}{6000}=0.08.
\displaystyle x=0.08\times6000=480.
\displaystyle \therefore \text{She bought }480\text{ tickets.}
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{If the probability of getting a bad egg in a lot of }400\text{ is }0.035,\text{ then the}
\displaystyle \text{number of bad eggs in the lot is }\underline{\hspace{2cm}}\text{.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the number of bad eggs}=x.
\displaystyle \frac{x}{400}=0.035.
\displaystyle x=0.035\times400=14.
\displaystyle \therefore \text{The number of bad eggs is }14.
\displaystyle \\

\displaystyle \textbf{Question 26: }\text{If an event cannot occur, i.e. it is an impossible event, then its probability is}
\displaystyle \underline{\hspace{2cm}}\text{.}
\displaystyle \text{Answer:}
\displaystyle P(\text{impossible event})=0.
\displaystyle \therefore \text{The required answer is }0.
\displaystyle \\

\displaystyle \textbf{Question 27: }\text{The probability that the month of April has exactly }5\text{ Mondays is}
\displaystyle \underline{\hspace{2cm}}\text{.}
\displaystyle \text{Answer:}
\displaystyle \text{April has }30\text{ days}=4\text{ weeks}+2\text{ days}.
\displaystyle \text{Thus, two consecutive weekdays occur }5\text{ times.}
\displaystyle \text{For Monday to occur }5\text{ times, the two extra days must be Sun-Mon or Mon-Tue.}
\displaystyle \therefore \text{Favourable number of elementary events}=2.
\displaystyle \text{Total number of elementary events}=7.
\displaystyle \therefore P(\text{exactly }5\text{ Mondays})=\frac{2}{7}.
\displaystyle \\

\displaystyle \textbf{Question 28: }\text{Two numbers }a\text{ and }b\text{ are selected, successively without replacement in}
\displaystyle \text{that order, from integers }1\text{ to }10.\text{ The probability that }\frac{a}{b}\text{ is an integer, is }\underline{\hspace{2cm}}\text{.}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of ordered outcomes}=10\times9=90.
\displaystyle \text{For }b=1,\text{ possible values of }a\text{ are }2,3,4,5,6,7,8,9,10\text{, giving }9\text{ outcomes.}
\displaystyle \text{For }b=2,\text{ possible values of }a\text{ are }4,6,8,10\text{, giving }4\text{ outcomes.}
\displaystyle \text{For }b=3,\text{ possible values of }a\text{ are }6,9\text{, giving }2\text{ outcomes.}
\displaystyle \text{For }b=4\text{ and }5,\text{ possible values of }a\text{ are }8\text{ and }10\text{ respectively.}
\displaystyle \therefore \text{Number of favourable outcomes}=9+4+2+1+1=17.
\displaystyle \therefore P\left(\frac{a}{b}\text{ is an integer}\right)=\frac{17}{90}.
\displaystyle \\

\displaystyle \textbf{Question 29: }\text{A box contains }5\text{ apples, }6\text{ oranges and }x\text{ bananas. If the probability of}
\displaystyle \text{selecting an apple from the box is }\frac{1}{3},\text{ then the number of bananas is }\underline{\hspace{2cm}}\text{.}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of fruits}=5+6+x=11+x.
\displaystyle P(\text{an apple})=\frac{5}{11+x}.
\displaystyle \therefore \frac{5}{11+x}=\frac{1}{3}.
\displaystyle 15=11+x.
\displaystyle \therefore x=4.
\displaystyle \\

\displaystyle \textbf{Question 30: }\text{A die has its six faces marked }0,1,1,1,6,6.\text{ If two such dice are thrown}
\displaystyle \text{together, then the probability of getting a total of }7\text{ is }\underline{\hspace{2cm}}\text{.}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of elementary events}=6\times6=36.
\displaystyle \text{A total of }7\text{ is obtained as }(1,6)\text{ or }(6,1).
\displaystyle \text{Number of ways of getting }(1,6)=3\times2=6.
\displaystyle \text{Number of ways of getting }(6,1)=2\times3=6.
\displaystyle \therefore \text{Number of favourable outcomes}=6+6=12.
\displaystyle \therefore P(\text{a total of }7)=\frac{12}{36}=\frac{1}{3}.
\displaystyle \\

\displaystyle \textbf{Question 31: }\text{A letter of English alphabets is chosen at random. The probability that it is a}
\displaystyle \text{consonant is }\underline{\hspace{2cm}}\text{.}
\displaystyle \text{Answer:}
\displaystyle \text{Number of letters in the English alphabet}=26.
\displaystyle \text{Number of vowels}=5.
\displaystyle \therefore \text{Number of consonants}=26-5=21.
\displaystyle \therefore P(\text{a consonant})=\frac{21}{26}.
\displaystyle \\

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