Probability Questions and Answers

Answer: A

Total number of balls = 15

Number of balls that are either black or white = 8

Hence the number of favourable outcomes of ball drawn being black or white is 8.

We know that, Probability of an event E, P (E) =number of favourable outcomes/total number of outcomes

So, the required probability is 8/15.

Answer: D

Given,

Number of green balls = 5

Number of black balls = 6

Number of white balls = 7

Total number of outcomes = 5 + 6 + 7 = 18

There are 18 balls out of which 11 are not white.

⇒ Number of favourable outcomes = 11

Probability of an event, P (E) = =Number of favourable outcomes / Total number of outcomes

⇒ P (ball drawn is not white) = 11/18

Therefore, probability that the ball drawn is not white is 11/18.

Alternate Method:

P (ball drawn is white) = 7/18

We know that,

P (ball drawn is white) + P (ball drawn is not white) = 1

Because the sum of the probability of an event and its complementary event is always 1.

⇒ P (ball drawn is not white) = 1 - P (ball drawn is white) = 1−7/18 = 11/18

Answer: B

A pack has 13 cards in each of the four suits.

\[ P = \frac{13}{52} = \frac{1}{4} \]

The four suits are equally represented, so each has probability \( \tfrac{1}{4} \) without any counting.

Answer: A

There are 4 jacks and 13 spades, but the jack of spades is in both groups and must not be counted twice.

\[ P(J \cup S) = P(J) + P(S) - P(J \cap S) \]

\[ = \frac{4}{52} + \frac{13}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13} \]

Adding without the correction gives \( \tfrac{17}{52} \) - the last option.

Answer: A

Let E be the event of getting different values on both the dice. The complementary event is getting the same value on both, for which there are 6 favorable outcomes: (1,1), (2,2), (3,3), (4,4), (5,5) and (6,6). Thus, P (not E) = 1/6

Thus, P (E) = 1 - P (not E) = 1 - (1/6) = 5/6

Answer: B

An event and its complement are exhaustive: between them they cover every possible outcome, and they cannot both happen.

\[ P(E) + P(\bar{E}) = 1 \]

This is why \( P(\bar{E}) = 1 - P(E) \), the form used whenever the complement is easier to count than the event itself.

Answer: D

There are only two colours of sock in the drawer, so consider the worst case.

The first two socks might be one white and one grey. But the third must match one of them, since there is no third colour available.

\[ P = 1 \]

This is the pigeonhole principle: three items sorted into two categories force at least two into the same one. The actual numbers of pairs never enter the argument.

Answer: B

There are two ways this can happen: even on the first and odd on the second, or the other way round.

\[ P = \frac{3}{6} \times \frac{3}{6} + \frac{3}{6} \times \frac{3}{6} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \]

Counting only one arrangement gives \( \tfrac{1}{4} \) - the first option - but the question does not say which die shows which.

Answer: B

The pot holds \( 2 + 6 + 4 + 8 = 20 \) balls.

Black and green are mutually exclusive, so their probabilities add:

\[ P = \frac{6}{20} + \frac{8}{20} = \frac{14}{20} = \frac{7}{10} \]

No ball is both black and green, so no overlap has to be subtracted.

Answer: D

Choose one of the 13 spades and one of the 13 hearts.

\[ P = \frac{13 \times 13}{{}^{52}C_{2}} = \frac{169}{1326} = \frac{13}{102} \]

The two cards are drawn together, so the pair is unordered and the denominator uses a combination; the numerator needs no factor of 2, since choosing one from each suit already counts each pair once.