Practice MCQ Questions and Answer on Probability

61.

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) $$\frac{1}{2}$$

Show Answer

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

62.

A box contains 10 black and 10 white balls. What is the probability of drawing 2 balls of the same colour ?

• (A) $$\frac{9}{19}$$
• (B) $$\frac{9}{38}$$
• (C) $$\frac{10}{19}$$
• (D) $$\frac{5}{19}$$

Show Answer

 Total number of balls = (10 + 10) = 20 Let E be the event of drawing 2 balls of the same colour. n (E) = number of ways of drawing 2 black balls or 2 white balls. n (E) =   $$\left( {{}^{10}\mathop C\nolimits_2 \times {}^{10}\mathop C\nolimits_2 } \right)$$   $$ = 2 \times {}^{10}\mathop C\nolimits_2 $$   $$ = 2 \times \frac{{10 \times 9}}{{2 \times 1}}$$   = 90 n (S) = number of ways of drawing 2 balls out of 20 balls $$ = {}^{20}\mathop C\nolimits_2 = \frac{{20 \times 19}}{{2 \times 1}}$$   = 190 $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{90}}{{190}} = \frac{9}{{19}}$$

63.

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) $$\frac{{5}}{{7}}$$

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 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 $$\eqalign{ & {\text{n}}\left( {\text{E}} \right)\, = {}^5{C_2} \cr & {\text{n}}\left( {\text{E}} \right) = \frac{{\left( {5 \times 4} \right)}}{{\left( {2 \times 1} \right)}} \cr & {\text{n}}\left( {\text{E}} \right) = 10 \cr & \therefore {\text{P}}\left( {\text{E}} \right) = \frac{{{\text{n}}\left( {\text{E}} \right)}}{{{\text{n}}\left( {\text{S}} \right)}} = \frac{{10}}{{21}} \cr} $$

64.

An urn contains 6 red, 5 blue and 2 green marbles. If 2 marbles are picked at random, what is the probability that both are red?

• (A) $$\frac{6}{{13}}$$
• (B) $$\frac{{5}}{{26}}$$
• (C) $$\frac{{5}}{{13}}$$
• (D) $$\frac{{7}}{{26}}$$

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 P(Both are red), $$\eqalign{ & = \frac{{^6{C_2}}}{{^{13}{C_2}}} \cr & = \frac{5}{{26}} \cr} $$

65.

Two cards are drawn from a pack of well shuffled cards. Find the probability that one is a club and other in King.

• (A) $$\frac{{1}}{{13}}$$
• (B) $$\frac{{4}}{{13}}$$
• (C) $$\frac{{1}}{{52}}$$
• (D) $$\frac{{1}}{{26}}$$

Show Answer

 Let X be the event that cards are in a club which is not king and other is the king of club. Let Y be the event that one is any club card and other is a non-club king. Hence, required probability: $$\eqalign{ & = P(A) + P(B) \cr & = \frac{{^{12}{C_1}{ \times ^1}{C_1}}}{{^{52}{C_2}}} + \frac{{^{13}{C_1}{ \times ^3}{C_1}}}{{^{52}{C_2}}} \cr & = \left( {\frac{{2 \times \left( {12 \times 1} \right)}}{{52 \times 51}}} \right) + \left( {\frac{{2\left( {13 \times 3} \right)}}{{52 \times 51}}} \right) \cr & = \left( {\frac{{24 + 78}}{{52 \times 51}}} \right) \cr & = \frac{1}{{26}} \cr} $$

66.

In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that the selected students are 2 boys and 1 girls, is:

• (A) $$\frac{21}{46}$$
• (B) $$\frac{25}{117}$$
• (C) $$\frac{1}{50}$$
• (D) $$\frac{3}{25}$$

Show Answer

 Let S be the sample space and let E be the event of selecting 2 boys and 1 girl. Then, n(S) = number of ways of selecting 3 students out of 25 = $${}^{25}\mathop C\nolimits_3 = $$   $$\frac{{25 \times 24 \times 23}}{{3 \times 2 \times 1}}$$   = 2300 And, n(E) = $$\left( {{}^{15}\mathop C\nolimits_2 \times {}^{10}\mathop C\nolimits_1 } \right)$$   $$ = \left( {\frac{{15 \times 14}}{{2 \times 1}} \times 10} \right)$$    = 1050 $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{1050}}{{2300}} = \frac{{21}}{{46}}$$

67.

An urn contains 6 red, 4 blue, 2 green and 3 yellow marbles. If three marbles are picked up at random, what is the probability that 2 are blue and 1 is yellow ?

• (A) $$\frac{{13}}{{35}}$$
• (B) $$\frac{{1}}{{5}}$$
• (C) $$\frac{{18}}{{455}}$$
• (D) $$\frac{{7}}{{15}}$$

Show Answer

 Total number of marbles = (6 + 4 + 2 + 3) = 15 Let E be the event of drawing 2 blue and 1 yellow marble. Then, n(E) = $$\left( {{}^4\mathop C\nolimits_2 \times {}^3\mathop C\nolimits_1 } \right)$$   $$ = \left( {\frac{{4 \times 3}}{{2 \times 1}} \times 3} \right)$$   = 18 Also, n(S) = $${}^{15}\mathop C\nolimits_3 = $$   $$\frac{{15 \times 14 \times 13}}{{3 \times 2 \times 1}}$$   = 455 $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{18}}{{455}}$$

68.

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 $$\eqalign{ & n\left( S \right) = {}^{52}{C_2} = \frac{{ {52 \times 51} }}{{ {2 \times 1} }} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1326 \cr} $$ 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 $$\eqalign{ & {\kern 1pt} = {^{13}{C_1}{ \times ^{13}}{C_1}} \cr & {\kern 1pt} {\kern 1pt} = {13 \times 13} {\kern 1pt} {\kern 1pt} \cr & {\kern 1pt} = 169 \cr & \therefore P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} = \frac{{169}}{{1326}} = \frac{{13}}{{102}} \cr} $$

69.

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) $$\frac{{5}}{{16}}$$

Show Answer

 In a simultaneous throw of two dice, we have n(S) = (6 x 6) = 36 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)} $$\eqalign{ & \therefore n\left( E \right) = 27 \cr & \therefore P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} = \frac{{27}}{{36}} = \frac{3}{4} \cr} $$

70.

In a simultaneous throw of two dice, what is the probability of getting a total of 7?

• (A) $$\frac{1}{6}$$
• (B) $$\frac{1}{4}$$
• (C) $$\frac{2}{3}$$
• (D) $$\frac{3}{4}$$

Show Answer

 We know that in a simultaneous throw of two dice, n(S) = 6 × 6 = 36 Let E = event of getting a total of 7 = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)} $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{6}{{36}} = \frac{1}{6}$$