Practice MCQ Questions and Answer on Probability

31.

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) $$\frac{{3}}{{7}}$$

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 Total number of balls = (6 + 8) = 14 Number of white balls = 8 P (drawing a white ball) = $$\frac{{8}}{{14}}$$  = $$\frac{{4}}{{7}}$$

32.

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

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

33.

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) $$\frac{{4}}{{9}}$$

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 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 $$\eqalign{ & P\left( A \right) = \frac{{n\left( A \right)}}{{n\left( S \right)}} \cr & = \frac{{10!\, \times 2!}}{{11!}} \cr & = \frac{2}{{11}} \cr} $$

34.

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

• (A) $$\frac{3}{10}$$
• (B) $$\frac{3}{20}$$
• (C) $$\frac{2}{5}$$
• (D) $$\frac{1}{2}$$

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

35.

A bag contains 4 red balls, 6 blue balls and 8 pink balls. One ball is drawn at random and replaced with 3 pink balls. A probability that the first ball drawn was either red or blue in colour and the second drawn was pink in colour ?

• (A) $$\frac{12}{21}$$
• (B) $$\frac{13}{17}$$
• (C) $$\frac{11}{30}$$
• (D) $$\frac{13}{18}$$

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Answer & Solution Answer: Option E Solution: Number of Red balls = 4 Number of Blue balls = 6 Number of Pink balls = 8 Total number of balls = 4 + 6 + 8 = 18 Required probability $$\eqalign{ & = \frac{4}{{18}} \times \frac{{11}}{{20}} + \frac{6}{{18}} \times \frac{{11}}{{20}} \cr & = \frac{{11}}{{20}}\left[ {\frac{4}{{18}} + \frac{6}{{18}}} \right] \cr & = \frac{{11}}{{20}} \times \frac{{10}}{{18}} \cr & = \frac{{11}}{{36}} \cr} $$

36.

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 $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{220}}{{455}} = \frac{{44}}{{91}}$$

37.

Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even ?

• (A) $$\frac{1}{2}$$
• (B) $$\frac{3}{4}$$
• (C) $$\frac{3}{8}$$
• (D) $$\frac{5}{16}$$

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 In a simultaneous throw of two dice, we have n (S) = (6 × 6) = 36 Let E = event of getting two numbers whose product is even. Then, E = {(1, 2), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (3, 4), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 2), (5, 4), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)} ∴ n (E) = 27 $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{27}}{{36}} = \frac{3}{4}$$

38.

Two cards are drawn together from a pack of 52 cards. The probability that one is a spade and one is a heart, is-

• (A) $$\frac{{3}}{{20}}$$
• (B) $$\frac{{29}}{{34}}$$
• (C) $$\frac{{47}}{{100}}$$
• (D) $$\frac{{13}}{{102}}$$

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 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 1 spade and 1 heart. ∴ n(E) = number of ways of choosing 1 spade out of 13 and 1 heart out of 13 $$ = \left( {{}^{13}\mathop C\nolimits_1 \times {}^{13}\mathop C\nolimits_1 } \right)$$   = (13 × 13) = 169 $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{169}}{{1326}} = \frac{{13}}{{102}}$$

39.

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) $$\frac{4}{105}$$

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 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 $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{4}{{105}}$$

40.

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) $$\frac{{9}}{{52}}$$

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