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 red nor green?

• (A) $$\frac{{1}}{{3}}$$
• (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 red nor green = event that the ball drawn is blue $$\eqalign{ & \therefore n(E) = 7 \cr & \therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{7}{{21}} = \frac{1}{3} \cr} $$

2.

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

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

3.

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

 Getting at most Two heads means 0 to 2 but not more than 2 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} $$\therefore P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} = \frac{7}{8}$$

4.

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

• (A) $$\frac{7}{12}$$
• (B) $$\frac{37}{44}$$
• (C) $$\frac{5}{12}$$
• (D) $$\frac{7}{44}$$

Show Answer

 Total number of marbles = (4 + 5 + 3) = 12 Let E be the event of drawing 3 marbles such that none is blue. Then, n (E) = number of ways of drawing 3 marbles out of 7 = $${}^7\mathop C\nolimits_3 $$   $$ = \frac{{7 \times 6 \times 5}}{{3 \times 2 \times 1}}$$   = 35 And, $$n(S) = {}^{12}\mathop C\nolimits_3 $$   $$ = \frac{{12 \times 11 \times 10}}{{3 \times 2 \times 1}}$$   = 220 $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{35}}{{220}} = \frac{7}{{44}}$$ ∴ Required probability = 1 - P(E) = $$\left( {1 - \frac{7}{{44}}} \right)$$ = $$ \frac{{37}}{{44}}$$

5.

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) $$\frac{27}{52}$$

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

6.

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 $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{15}}{{105}} = \frac{1}{7}$$

7.

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 × 6) = 36 Let E = event of getting two numbers whose product is even. 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)} ∴ n (E) = 27 $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{27}}{{36}} = \frac{3}{4}$$

8.

A basket contains 4 red, 5 blue and 3 green marbles. If 2 marbles are drawn at random from the basket, What is the probability that both are red ?

• (A) $$\frac{{3}}{{7}}$$
• (B) $$\frac{{1}}{{2}}$$
• (C) $$\frac{{1}}{{11}}$$
• (D) $$\frac{{1}}{{6}}$$

Show Answer

 Total number of balls = (4 + 5 + 3) = 12 Let E be the event of drawing 2 red balls. Then, n (E) = $${}^4\mathop C\nolimits_2 = \frac{{4 \times 3}}{{2 \times 1}}$$   = 6 Also n (S) = $${}^{12}\mathop C\nolimits_2 = \frac{{12 \times 11}}{{2 \times 1}}$$   = 66 $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{6}{{66}} = \frac{1}{{11}}$$

9.

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 $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{90}}{{190}} = \frac{9}{{19}}$$

10.

A bag contains 6 red balls 11 yellow balls and 5 pink balls. If two balls are drawn at random from the bag. One after another what is the probability that the first ball is red and second ball is yellow?

• (A) $$\frac{1}{14}$$
• (B) $$\frac{2}{7}$$
• (C) $$\frac{1}{7}$$
• (D) $$\frac{3}{14}$$

Show Answer

 Number of red balls = 6 Number of yellow balls = 11 Number of pink balls = 5 Total number of balls = 6 + 11 + 5 = 22 Total possible outcomes $$n(E) = {}^{22}\mathop C\nolimits_2 = \frac{{22!}}{{2!(22 - 2)!}}$$ $$ = \frac{{22!}}{{2! \times 20!}} $$ $$ = \frac{{22 \times 21}}{{2 \times 1}} $$ $$=$$ 231 Number of favourable outcomes $$n(S) = {}^6\mathop C\nolimits_1 \times {}^{11}\mathop C\nolimits_1 $$    = 6 × 11 = 66 Required probability = $$\frac{{n(E)}}{{n(S)}}$$   $$ = \frac{{66}}{{231}}$$   $$ = \frac{1}{7}$$