Practice MCQ Questions and Answer on Probability

1.

In a box, there are 8 red, 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither blue nor green ?

• (A) $$\frac{2}{5}$$
• (B) $$\frac{3}{4}$$
• (C) $$\frac{7}{19}$$
• (D) $$\frac{8}{21}$$

Show Answer

 Total number of balls = (8 + 7 + 6) = 21 Let E = event that the ball drawn is neither blue nor green = Even that the ball drawn is red ∴ n(E) = 8 P(E) = $$\frac{{8}}{{21}}$$

2.

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 $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{1050}}{{2300}} = \frac{{21}}{{46}}$$

3.

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)} $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{6}{{36}} = \frac{1}{6}$$

4.

What is the probability of getting a sum 9 from two throws of a dice ?

• (A) $$\frac{1}{6}$$
• (B) $$\frac{1}{8}$$
• (C) $$\frac{1}{9}$$
• (D) $$\frac{1}{12}$$

Show Answer

 In two throws of dice, n(S) = (6 × 6) = 36 Let E = event of getting a sum 9 = [(3, 6), (4, 5), (5, 4), (6, 3)] $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{4}{{36}} = \frac{1}{9}$$

5.

An urn contains 6 red, 4 blue, 2 green 3 yellow marbles. If two marbles are drawn at random from the run, what is the probability that both are red ?

• (A) $$\frac{1}{6}$$
• (B) $$\frac{1}{7}$$
• (C) $$\frac{2}{15}$$
• (D) $$\frac{2}{5}$$

Show Answer

 Total number of balls = (6 + 4 + 2 + 3) = 15 Let E be the event of drawing 2 red balls. Then, n(E)  $$ = {}^6\mathop C\nolimits_2 $$  $$ = \frac{{6 \times 5}}{{2 \times 1}}$$   = 15 Also, $$n(S) = {}^{15}\mathop C\nolimits_2 $$   $$ = \frac{{15 \times 14}}{{2 \times 1}}$$   = 105 $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{15}}{{105}} = \frac{1}{7}$$

6.

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

7.

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 $$\eqalign{ & \therefore P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} \cr & = \frac{2}{{52}} \cr & = \frac{1}{{26}} \cr} $$

8.

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

9.

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) $$\frac{13}{102}$$

Show Answer

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

10.

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) - $$P(E \cup H)$$ $$\eqalign{ & = \left( {\frac{3}{{10}} + \frac{1}{5} - \frac{1}{{10}}} \right) \cr & = \frac{4}{{10}} \cr & = \frac{2}{5} \cr} $$