Practice MCQ Questions and Answer on Probability

61.

Three unbiased coins are tossed. What is the probability of getting at most two heads?

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

Show Answer

 Here S = [TTT, TTH, THT, HTT, THH, HTH, HHT, HHH] Let E = event of getting at most two heads Then, E = {TTT, TTH, THT, HTT, THH, HTH, HHT} $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{7}{8}$$

62.

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

63.

In a single throw of die, what is the probability of getting a number greater than 4 ?

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

Show Answer

 When a die is thrown, we have S = {1, 2, 3, 4, 5, 6} Let, E = event of getting a number greater than 4 = {5, 6} $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{2}{6} = \frac{1}{3}$$

64.

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

65.

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 $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{2}{{52}} = \frac{1}{{26}}$$

66.

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 $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{28}}{{52}} = \frac{7}{{13}}$$

67.

One card is drawn at random from a pack of 52 cards. What is the probability that the card drawn is a face card ?

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

Show Answer

 Clearly, there are 52 cards, out of which there are 12 face cards 4 jack, 4 queens, and 4 kings ∴ P (getting a face card) $$ = \frac{{12}}{{52}}$$   $$ = \frac{3}{{13}}$$

68.

Two cards are drawn from a pack of 52 cards. The probability that either both are red or both are king, is-

• (A) $$\frac{7}{13}$$
• (B) $$\frac{3}{26}$$
• (C) $$\frac{63}{221}$$
• (D) $$\frac{55}{221}$$

Show Answer

 Clearly, n (S) = $$n{\text{ }}(S) = $$   $${}^{52}\mathop C\nolimits_2 = $$   $$\frac{{\left( {52 \times 51} \right)}}{2}$$   = 1326 Let $${{E_1}}$$ = event of getting both red cards $${{E_2}}$$ = event of getting both kings Then, $${{E_1}}$$ $$ \cap $$ $${{E_2}}$$ = event of getting 2 kings of red cards. ∴ $$n{\text{ }}({E_1}) = {}^{26}\mathop C\nolimits_2 = \frac{{\left( {26 \times 25} \right)}}{{\left( {2 \times 1} \right)}}$$     = 325 and $$n{\text{ }}({E_2}) = {}^4\mathop C\nolimits_2 = \frac{{\left( {4 \times 3} \right)}}{{\left( {2 \times 1} \right)}}$$     = 6 $$n\left( {{E_1} \cap {E_2}} \right) = {}^2{C_2} = 1$$ $$\therefore P({E_1}) = \frac{{n({E_1})}}{{n(S)}} = \frac{{325}}{{1326}}$$      and $$P({E_2}) = \frac{{n({E_2})}}{{n(S)}} = \frac{6}{{1326}}$$ $$P({E_1} \cap {E_2}) = \frac{1}{{1326}}$$ ∴ P (both red or both kings) $$\eqalign{ & = P\left( {{E_1} \cup {E_2}} \right) \cr & = P\left( {{E_1}} \right) + P\left( {{E_2}} \right) - P\left( {{E_1} \cap {E_2}} \right) \cr & = \left( {\frac{{325}}{{1326}} + \frac{6}{{1326}} - \frac{1}{{1326}}} \right) \cr & = \frac{{330}}{{1326}} \cr & = \frac{{55}}{{221}} \cr} $$

69.

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

70.

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

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