Practice MCQ Questions and Answer on Probability

51.

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

52.

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

53.

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

54.

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) $$\eqalign{ & = P\left( {{E_1} \cup {E_2}} \right) \cr & = P\left( {{E_1}} \right) + P\left( {{E_2}} \right) - P\left( {{E_1} \cap {E_2}} \right) \cr & = \left( {\frac{{325}}{{1326}} + \frac{6}{{1326}} - \frac{1}{{1326}}} \right) \cr & = \frac{{330}}{{1326}} \cr & = \frac{{55}}{{221}} \cr} $$

55.

In a class, 30% of the students offered English, 20% offered Hindi and 10% offered both. If a student is selected at random, What is the probability that he has offered English or Hindi ?

• (A) $$\frac{2}{5}$$
• (B) $$\frac{3}{5}$$
• (C) $$\frac{3}{4}$$
• (D) $$\frac{3}{10}$$

Show Answer

 $$\eqalign{ & P(E) = \frac{{30}}{{100}} = \frac{3}{{10}}; \cr & P(H) = \frac{{20}}{{100}} = \frac{1}{5}\,\,{\text{and}} \cr & {\text{P }}\left( {{\text{E}} \cap {\text{H}}} \right)\,{\text{ = }}\frac{{10}}{{100}} = \frac{1}{{10}} \cr} $$ P (E or H) = $$P(E \cup H)$$ = P(E) + P(H) - $$P(E \cup H)$$ $$\eqalign{ & = \left( {\frac{3}{{10}} + \frac{1}{5} - \frac{1}{{10}}} \right) \cr & = \frac{4}{{10}} \cr & = \frac{2}{5} \cr} $$

56.

A basket contains 6 blue, 2 red, 4 green and 3 yellow balls. If 5 balls are picked up at random, what is the probability that at least one is blue?

• (A) $$\frac{137}{143}$$
• (B) $$\frac{18}{455}$$
• (C) $$\frac{9}{91}$$
• (D) $$\frac{2}{5}$$

Show Answer

 Total number of balls = (6 + 2 + 4 + 3) = 15 Let E be the event of drawing 5 balls out of 9 non-blue balls. $$ = {}^9\mathop C\nolimits_5 = {}^9\mathop C\nolimits_{\left( {9 - 5} \right)} = {}^9\mathop C\nolimits_4 $$    $$ = \frac{{9 \times 8 \times 7 \times 6}}{{4 \times 3 \times 2 \times 1}}$$    = 126 And, n(S) = $${}^{15}\mathop C\nolimits_5 = $$   $$\frac{{15 \times 14 \times 13 \times 12 \times 11}}{{5 \times 4 \times 3 \times 2 \times 1}}$$     = 3003 $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = $$    $$\frac{{126}}{{3003}} = \frac{6}{{143}}$$ ∴ Required Probability $$\eqalign{ & = \left( {1 - \frac{6}{{143}}} \right) \cr & = \frac{{137}}{{143}} \cr} $$

57.

The probability that a card drawn from a pack of 52 cards will be a diamond or a king is -

• (A) $$\frac{2}{13}$$
• (B) $$\frac{4}{13}$$
• (C) $$\frac{1}{13}$$
• (D) $$\frac{1}{52}$$

Show Answer

 Here, n(S) = 52 There are 13 cards of diamond (including one king) and there are three more kings. Let E = event of getting a diamond or a king Then, n(E) = (13 + 3) = 16 $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{16}}{{52}} = \frac{4}{{13}}$$

58.

A bag contains 4 white, 5 red and 6 blue balls. Three balls are drawn at random from the bag. The probability that all of them are red, is:

• (A) $$\frac{1}{22}$$
• (B) $$\frac{3}{22}$$
• (C) $$\frac{2}{91}$$
• (D) $$\frac{2}{77}$$

Show Answer

 Let S be the sample space Then, n(S) = number of ways of drawing 3 balls out of 15 $$\eqalign{ & = {}^{15}{C_3} \cr & = \frac{{ {15 \times 14 \times 13} }}{{ {3 \times 2 \times 1} }} \cr & = 455 \cr} $$ Let E = event of getting all the 3 red balls $$\eqalign{ & \therefore n\left( E \right) = {}^5{C_3} = {}^5{C_2} \cr & = \frac{{ {5 \times 4} }}{{ {2 \times 1} }} = 10 \cr & \therefore P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} \cr & = \frac{{10}}{{455}} \cr & = \frac{2}{{91}} \cr} $$

59.

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

 Here, n(S) = 52 Let E = event of getting a queen of club or a king of heart Then, n(E) = 2 $$\eqalign{ & \therefore P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} \cr & = \frac{2}{{52}} \cr & = \frac{1}{{26}} \cr} $$

60.

In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected, 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 E be the event of selecting 1 girl and 2 boys Then, n(S) = Number ways of selecting 3 students out of 25 $$\eqalign{ & = {}^{25}{C_3} \cr & = \frac{{ {25 \times 24 \times 23} }}{{ {3 \times 2 \times 1} }} \cr & = 2300 \cr & n\left( E \right) = {^{10}{C_1}{ \times ^{15}}{C_2}} \cr & = {10 \times \frac{{ {15 \times 14} }}{{ {2 \times 1} }}} \cr & = 1050 \cr & \therefore P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} = \frac{{1050}}{{2300}} = \frac{{21}}{{46}} \cr} $$