Practice MCQ Questions and Answer on Probability

41.

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} $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{9}{{20}}$$

42.

Three unbiased coins are tossed. What is the probability of getting at least 2 heads?

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

Show Answer

 Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH} Let E = event of getting at least two heads = {THH, HTH, HHT, HHH} $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{4}{8} = \frac{1}{2}$$

43.

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} $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{6}{{20}} = \frac{3}{{10}}$$

44.

A box contains 4 red, 5 green and 6 white balls. A ball is drawn at random from the box. What is the probability that the ball drawn is either red or green ?

• (A) $$\frac{2}{5}$$
• (B) $$\frac{3}{5}$$
• (C) $$\frac{1}{5}$$
• (D) $$\frac{7}{15}$$

Show Answer

 Total number of balls = (4 + 5 + 6) = 15 P (drawing a red ball or a green ball) = P (red) + P (green) $$\eqalign{ & = \left( {\frac{4}{{15}} + \frac{5}{{15}}} \right) \cr & = \frac{9}{{15}} \cr & = \frac{3}{5} \cr} $$

45.

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)] $$\eqalign{ & = P\left[ {\left( {{E_1} \cap {{\overline E }_2}} \right)or\left( {{{\overline E }_1} \cap {E_2}} \right)} \right] \cr & = P\left[ {\left( {{E_1} \cap {{\overline E }_2}} \right) + \left( {{{\overline E }_1} \cap {E_2}} \right)} \right] \cr & = P\left( {{E_1}} \right).P\left( {{{\overline E }_2}} \right) + P\left( {{{\overline E }_1}} \right).P\left( {{E_2}} \right) \cr & = \left( {\frac{3}{4} \times \frac{1}{5}} \right) + \left( {\frac{1}{4} \times \frac{4}{5}} \right) \cr & = \left( {\frac{3}{{20}} + \frac{1}{5}} \right) \cr & = \frac{7}{{20}} \cr & = \left( {\frac{7}{{20}} \times 100} \right)\% \cr & = 35\% \cr} $$

46.

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} $$\therefore P\left( E \right) = \frac{{n(E)}}{{n(S)}} = \frac{3}{4}$$

47.

From a pack of 52 cards, one card is drawn at random. What is the probability that the card drawn is a ten or a spade?

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

Show Answer

 Hence, n (S) = 52 There are 13 spades (including one ten) and there are 3 more ten Let E = event of getting a ten or a spade Then, n (E) = (13 + 3) = 16 $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{16}}{{52}} = \frac{4}{{13}}$$

48.

One card is drawn from a pack of 52 cards. What is the probability that the card drawn is either a red card or a king ?

• (A) $$\frac{1}{2}$$
• (B) $$\frac{6}{13}$$
• (C) $$\frac{7}{13}$$
• (D) $$\frac{27}{52}$$

Show Answer

 Here, n(S) = 52 There are 26 red cards (including 2 kings) and there are 2 more kings. Let E = event of getting a red card or a king. Then, n(E) = 28 $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{28}}{{52}} = \frac{7}{{13}}$$

49.

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 $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{15}}{{36}} = \frac{5}{{12}}$$

50.

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} $$\therefore P\left( E \right) = \frac{{n(E)}}{{n(S)}} = \frac{3}{4}$$