Practice MCQ Questions and Answer on Probability

11.

The probability that a card drawn from a pack of 52 cards will be a diamond or a king is -

• (A) $$\frac{2}{13}$$
• (B) $$\frac{4}{13}$$
• (C) $$\frac{1}{13}$$
• (D) $$\frac{1}{52}$$

Show Answer

 Here, n(S) = 52 There are 13 cards of diamond (including one king) and there are three more kings. Let E = event of getting a diamond or a king Then, n(E) = (13 + 3) = 16 $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{16}}{{52}} = \frac{4}{{13}}$$

12.

A speaks truth in 75% cases and B in 80% of the cases. In what percentage of cases are they likely to contradict each other, in narrating the same incident ?

• (A) 5%
• (B) 15%
• (C) 35%
• (D) 45%

Show Answer

 Let $${{{\text{E}}_1}}$$ = event that A speaks the truth And $${{{\text{E}}_2}}$$ = event that B speaks the truth Then, $$\eqalign{ & P\left( {{E_1}} \right) = \frac{{75}}{{100}} = \frac{3}{4}, \cr & P\left( {{E_2}} \right) = \frac{{80}}{{100}} = \frac{4}{5}, \cr & P\left( {{{\overline E }_1}} \right) = \left( {1 - \frac{3}{4}} \right) = \frac{1}{4}, \cr & P\left( {{{\overline E }_2}} \right) = \left( {1 - \frac{4}{5}} \right) = \frac{1}{5} \cr} $$ P (A and B contradict each other) = P [(A speaks the truth and B tells a lie) or (A tells a lie and B speaks the truth)] $$\eqalign{ & = P\left[ {\left( {{E_1} \cap {{\overline E }_2}} \right)or\left( {{{\overline E }_1} \cap {E_2}} \right)} \right] \cr & = P\left[ {\left( {{E_1} \cap {{\overline E }_2}} \right) + \left( {{{\overline E }_1} \cap {E_2}} \right)} \right] \cr & = P\left( {{E_1}} \right).P\left( {{{\overline E }_2}} \right) + P\left( {{{\overline E }_1}} \right).P\left( {{E_2}} \right) \cr & = \left( {\frac{3}{4} \times \frac{1}{5}} \right) + \left( {\frac{1}{4} \times \frac{4}{5}} \right) \cr & = \left( {\frac{3}{{20}} + \frac{1}{5}} \right) \cr & = \frac{7}{{20}} \cr & = \left( {\frac{7}{{20}} \times 100} \right)\% \cr & = 35\% \cr} $$

13.

A bag contains 6 red balls 11 yellow balls and 5 pink balls. If two balls are drawn at random from the bag. One after another what is the probability that the first ball is red and second ball is yellow?

• (A) $$\frac{1}{14}$$
• (B) $$\frac{2}{7}$$
• (C) $$\frac{1}{7}$$
• (D) $$\frac{3}{14}$$

Show Answer

 Number of red balls = 6 Number of yellow balls = 11 Number of pink balls = 5 Total number of balls = 6 + 11 + 5 = 22 Total possible outcomes $$n(E) = {}^{22}\mathop C\nolimits_2 = \frac{{22!}}{{2!(22 - 2)!}}$$ $$ = \frac{{22!}}{{2! \times 20!}} $$ $$ = \frac{{22 \times 21}}{{2 \times 1}} $$ $$=$$ 231 Number of favourable outcomes $$n(S) = {}^6\mathop C\nolimits_1 \times {}^{11}\mathop C\nolimits_1 $$    = 6 × 11 = 66 Required probability = $$\frac{{n(E)}}{{n(S)}}$$   $$ = \frac{{66}}{{231}}$$   $$ = \frac{1}{7}$$

14.

Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5?

• (A) $$\frac{1}{2}$$
• (B) $$\frac{2}{5}$$
• (C) $$\frac{8}{15}$$
• (D) $$\frac{9}{20}$$

Show Answer

 Here, S = {1, 2, 3, 4, ...., 19, 20} Let E = event of getting a multiple of 3 or 5= {3, 6 , 9, 12, 15, 18, 5, 10, 20} $$\therefore P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} = \frac{9}{{20}}$$

15.

A man and his wife appear in an interview for two vacancies in the same post. The probability two of husband’s selection is $$\frac{1}{7}$$ and the probability of wife’s selection is$$\frac{1}{5}$$. What is the probability that only one of them is selected?

• (A) $$\frac{4}{5}$$
• (B) $$\frac{2}{7}$$
• (C) $$\frac{4}{7}$$
• (D) $$\frac{8}{15}$$

Show Answer

 Let $${E_1}$$ = event that the husband is selected and $${E_2}$$ = event that the wife is selected. Then, $$P({E_1}) = \frac{1}{7}{\text{ }}$$   and $${\text{ }}P({E_2}) = \frac{1}{5}$$ ∴ $$P({\overline E _1}) = \left( {1 - \frac{1}{7}} \right)$$   $${\text{ = }}\frac{6}{7}$$ and $$P({\overline E _2}) = $$   $$\left( {1 - \frac{1}{5}} \right) = \frac{4}{5}$$ ∴ Required probability = P [( A and not B) or (B and not A)] = P $$\left[ {\left( {{E_1} \cap {{\bar E}_2}} \right)\,{\text{or}}\,\left( {{E_2} \cap {{\bar E}_1}} \right)} \right]$$ = P $$\left( {{E_1} \cap {{\overline E }_2}} \right) + {\text{ P}}\left( {{E_2} \cap {{\overline E }_1}} \right)$$= $${\text{P}}\left( {{E_1}} \right).{\text{ P}}\left( {{{\overline E }_2}} \right) + P\left( {{E_2}} \right).P\left( {{{\overline E }_1}} \right)$$ $$\eqalign{ & = \left( {\frac{1}{7} \times \frac{4}{5}} \right) + \left( {\frac{1}{5} \times \frac{6}{7}} \right) \cr & = \frac{10}{{35}} \cr & = \frac{2}{7} \cr} $$

16.

A bag contains 4 red, 5 yellow and 6 pink balls. Two balls are drawn at random. What is the probability that none of the balls drawn are yellow?

• (A) $$\frac{1}{7}$$
• (B) $$\frac{3}{7}$$
• (C) $$\frac{2}{7}$$
• (D) $$\frac{5}{14}$$

Show Answer

 Number of red balls = 4 Number of yellow balls = 5 Number of pink balls = 6 Total balls = 4 + 5 + 6 = 15 Total possible outcomes = selection of 2 balls out of 15 balls = $${}^{15}\mathop C\nolimits_2 $$   =$$\frac{{15!}}{{2!(15 - 2)!}}$$ =$$\frac{{15!}}{{2! × 13!}}$$ =$$\frac{{15 × 14}}{{1 × 2}}$$ =105 Total favourable outcomes = selection of 2 balls out of 4 orange and 6 pink balls. $$ = {}^{10}\mathop C\nolimits_2 $$ = $$\frac{{10!}}{{2!(10 - 2)!}}$$ = $$\frac{{10!}}{{2! × 8!}}$$ = $$\frac{{10 × 9}}{{1 × 2}}$$ = 45 ∴ Required probability = $$\frac{{45}}{{105}}$$ = $$\frac{{3}}{{7}}$$

17.

An urn contains 6 red, 4 blue, 2 green and 3 yellow marbles. If two marbles are picked up at random, what is the probability that either both are green or both are yellow ?

• (A) $$\frac{5}{91}$$
• (B) $$\frac{1}{35}$$
• (C) $$\frac{1}{3}$$
• (D) $$\frac{4}{105}$$

Show Answer

 Total number of marbles = (6 + 4 + 2 + 3) = 15 Let E be the event of drawing 2 marbles such that either both are green or both are yellow. Then, n (E) = $$\left( {{}^2\mathop C\nolimits_1 + {}^3\mathop C\nolimits_2 } \right)$$   $$ = \left( {1 + {}^3\mathop C\nolimits_1 } \right)$$   = (1 + 3) = 4And, n (S) = $${}^{15}\mathop C\nolimits_2 = $$   $$\frac{{15 \times 14}}{{2 \times 1}}$$   = 105 $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{4}{{105}}$$

18.

One card is drawn at random from a pack of 52 cards. What is the probability that the card drawn is a face card (Jack, Queen and King only)?

• (A) $$\frac{1}{13}$$
• (B) $$\frac{3}{13}$$
• (C) $$\frac{1}{4}$$
• (D) $$\frac{9}{52}$$

Show Answer

 Clearly, there are 52 cards, out of which there are 12 face cards. $$\eqalign{ & \therefore P\left( {{\text{getting}}\,{\text{a}}\,{\text{face}}\,{\text{card}}} \right) \cr & = \frac{{12}}{{52}} = \frac{3}{{13}} \cr} $$

19.

An urn contains 6 red, 4 blue, 2 green 3 yellow marbles. If two marbles are drawn at random from the run, what is the probability that both are red ?

• (A) $$\frac{1}{6}$$
• (B) $$\frac{1}{7}$$
• (C) $$\frac{2}{15}$$
• (D) $$\frac{2}{5}$$

Show Answer

 Total number of balls = (6 + 4 + 2 + 3) = 15 Let E be the event of drawing 2 red balls. Then, n(E)  $$ = {}^6\mathop C\nolimits_2 $$  $$ = \frac{{6 \times 5}}{{2 \times 1}}$$   = 15 Also, $$n(S) = {}^{15}\mathop C\nolimits_2 $$   $$ = \frac{{15 \times 14}}{{2 \times 1}}$$   = 105 $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{15}}{{105}} = \frac{1}{7}$$

20.

A basket contains 6 blue, 2 red, 4 green, and 3 yellow balls. If three balls are picked up at random, what is the probability that none is yellow?

• (A) $$\frac{3}{455}$$
• (B) $$\frac{1}{5}$$
• (C) $$\frac{4}{5}$$
• (D) $$\frac{44}{91}$$

Show Answer

 Total number of balls = (6 + 2 + 4 + 3) = 15 Let E be the event of drawing 3 non-yellow balls Then, n(E) = $${}^{12}\mathop C\nolimits_3 $$   $$ = \frac{{12 \times 11 \times 10}}{{3 \times 2 \times 1}}$$   = 220 Also, n(S) = $${}^{15}\mathop C\nolimits_3 $$   $$ = \frac{{15 \times 14 \times 13}}{{3 \times 2 \times 1}}$$   = 455 $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{220}}{{455}} = \frac{{44}}{{91}}$$