Probability Questions and Answers

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.

Answer: C

The bag holds only red and blue marbles, so a black one cannot be drawn at all.

\[ P(\text{black}) = \frac{\text{number of black marbles}}{\text{total marbles}} = \frac{0}{5} = 0 \]

An event that cannot happen is an impossible event, and its probability is exactly zero.

Answer: B

Every probability satisfies

\[ 0 \leq P(E) \leq 1 \]

with 0 for an impossible event and 1 for a certain one. Both bounds are attainable, so the interval is inclusive, and no probability can exceed 1.

Answer: C

A binomial experiment requires exactly three things: each trial has only two outcomes, the probability of success is the same on every trial, and the trials are independent.

The number of trials \( n \) may be any positive integer - even one - so "must have at least 3 trials" is not a condition.