Practice MCQ Questions and Answer on Probability

41.

Two dice are tossed. The probability that the total score is a prime number is-

• (A) $$\frac{1}{6}$$
• (B) $$\frac{1}{2}$$
• (C) $$\frac{5}{12}$$
• (D) $$\frac{7}{9}$$

Show Answer

 Clearly, n (S) = (6 × 6) = 36 Let E be the event that the sum is a prime number. Then, n (E) = {(1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3,), (5, 2), (5, 6), (6, 1), (6, 5)} ∴ n (E) = 15 $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{15}}{{36}} = \frac{5}{{12}}$$

42.

In a lottery, there are 10 prizes and 25 blanks. A lottery is drawn at random. What is the probability of getting a prize ?

• (A) $$\frac{1}{10}$$
• (B) $$\frac{2}{5}$$
• (C) $$\frac{2}{7}$$
• (D) $$\frac{5}{7}$$

Show Answer

 P (getting a prize) $$\eqalign{ & = \frac{{10}}{{\left( {10 + 25} \right)}} \cr & = \frac{{10}}{{35}} \cr & = \frac{2}{7} \cr} $$

43.

A bag contains 2 red, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?

• (A) $$\frac{10}{21}$$
• (B) $$\frac{11}{21}$$
• (C) $$\frac{2}{7}$$
• (D) $$\frac{5}{7}$$

Show Answer

 Total number of balls = (2 + 3 + 2) = 7 Let S be the sample space Then, n(S) = Number of ways of drawing 2 balls out of 7 $$\eqalign{ & {\text{n}}\left( {\text{S}} \right) = {}^7{C_2} \cr & {\text{n}}\left( {\text{S}} \right) = \frac{{\left( {7 \times 6} \right)}}{{\left( {2 \times 1} \right)}} \cr & {\text{n}}\left( {\text{S}} \right) = 21 \cr} $$ Let E = Event of 2 balls, none of which is blue ∴ n(E) = Number of ways of drawing 2 balls out of (2 + 3) balls $$\eqalign{ & {\text{n}}\left( {\text{E}} \right)\, = {}^5{C_2} \cr & {\text{n}}\left( {\text{E}} \right) = \frac{{\left( {5 \times 4} \right)}}{{\left( {2 \times 1} \right)}} \cr & {\text{n}}\left( {\text{E}} \right) = 10 \cr & \therefore {\text{P}}\left( {\text{E}} \right) = \frac{{{\text{n}}\left( {\text{E}} \right)}}{{{\text{n}}\left( {\text{S}} \right)}} = \frac{{10}}{{21}} \cr} $$

44.

A bag contains 6 black and 8 white balls. One ball is drawn at random. What is the probability that the ball drawn is white?

• (A) $$\frac{3}{4}$$
• (B) $$\frac{4}{7}$$
• (C) $$\frac{1}{8}$$
• (D) $$\frac{3}{7}$$

Show Answer

 $$\eqalign{ & {\text{Let}}\,{\text{number}}\,{\text{of}}\,{\text{balls}} \cr & = \left( {6 + 8} \right) \cr & = 14 \cr & {\text{Number}}\,{\text{of}}\,{\text{white}}\,{\text{balls}} = 8 \cr & P\left( {{\text{drawing}}\,{\text{a}}\,{\text{white ball}}} \right) \cr & = \frac{8}{{14}} = \frac{4}{7} \cr} $$

45.

In a simultaneous throw of two coins, the probability of getting at least one head is-

• (A) $$\frac{1}{2}$$
• (B) $$\frac{1}{3}$$
• (C) $$\frac{2}{3}$$
• (D) $$\frac{3}{4}$$

Show Answer

 Here S = {HH, HT, TH, TT} Let E = event of getting at least one head = {HT, TH, HH} $$\therefore P\left( E \right) = \frac{{n(E)}}{{n(S)}} = \frac{3}{4}$$

46.

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 $$\left( {{E_1} \cap {{\overline E }_2}} \right) + {\text{ P}}\left( {{E_2} \cap {{\overline E }_1}} \right)$$= $${\text{P}}\left( {{E_1}} \right).{\text{ P}}\left( {{{\overline E }_2}} \right) + P\left( {{E_2}} \right).P\left( {{{\overline E }_1}} \right)$$ $$\eqalign{ & = \left( {\frac{1}{7} \times \frac{4}{5}} \right) + \left( {\frac{1}{5} \times \frac{6}{7}} \right) \cr & = \frac{10}{{35}} \cr & = \frac{2}{7} \cr} $$

47.

An urn contains 2 red, 3 green and 2 blue balls. If 2 balls are drawn at random, find the probability that no ball is blue.

• (A) $$\frac{5}{7}$$
• (B) $$\frac{10}{21}$$
• (C) $$\frac{2}{7}$$
• (D) $$\frac{11}{21}$$

Show Answer

 Total number of balls = (2 + 3 + 2) = 7 Let, E be the event of drawing 2 non-blue balls. Then, n (E) = $${}^5\mathop C\nolimits_4 = \frac{{5 \times 4}}{{2 \times 1}}$$   = 10 And, n (S) = $${}^7\mathop C\nolimits_2 = \frac{{7 \times 6}}{{2 \times 1}}$$   = 21 $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{10}}{{21}}$$

48.

P and Q sit in a ring arrangement with 10 persons. What is the probability that P and Q will sit together?

• (A) $$\frac{2}{11}$$
• (B) $$\frac{3}{11}$$
• (C) $$\frac{2}{9}$$
• (D) $$\frac{4}{9}$$

Show Answer

 n(S)= number of ways of sitting 12 persons at round table: = (12 - 1)! = 11! Since two persons will be always together, then number of persons: = 10 + 1 = 11 So, 11 persons will be seated in (11 - 1)! = 10! ways at round table and 2 particular persons will be seated in 2! ways. n(A) = The number of ways in which two persons always sit together = 10! × 2 $$\eqalign{ & P\left( A \right) = \frac{{n\left( A \right)}}{{n\left( S \right)}} \cr & = \frac{{10!\, \times 2!}}{{11!}} \cr & = \frac{2}{{11}} \cr} $$

49.

A card is drawn from a pack of 52 cards. The probability of getting a queen of club or a king of heart is-

• (A) $$\frac{1}{13}$$
• (B) $$\frac{2}{13}$$
• (C) $$\frac{1}{26}$$
• (D) $$\frac{1}{52}$$

Show Answer

 Hence, n(S) = 52 Let E = event of getting a queen of club or a king of heart. Then, n(E) = 2 $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{2}{{52}} = \frac{1}{{26}}$$

50.

From a pack of 52 cards, two cards are drawn together at random. What is the probability of both the cards being kings?

• (A) $$\frac{1}{15}$$
• (B) $$\frac{25}{57}$$
• (C) $$\frac{35}{256}$$
• (D) $$\frac{1}{221}$$

Show Answer

 Let S be the sample space Then, $$n(S) = $$ $${}^{52}\mathop C\nolimits_2 = $$ $$\frac{{\left( {52 \times 51} \right)}}{{\left( {2 \times 1} \right)}}$$   =1326 Let E = event of getting 2 kings out of 4 ∴ $$n(E) = {}^4\mathop C\nolimits_2 = $$   $$\frac{{\left( {4 \times 3} \right)}}{{\left( {2 \times 1} \right)}}$$   = 6 ∴ $$P(E) = \frac{{n(E)}}{{n(S)}} = $$   $$\frac{6}{{1326}} = \frac{1}{{221}}$$