Probability Questions and Answers

Answer: A

The probability of a red card is 1/2.

Half the deck is red: 13 hearts and 13 diamonds, 26 of the 52 cards. So P(red) = 26/52 = 1/2.

1/4 is the probability of one particular suit, 1/13 of one particular rank, and 1/52 of one particular card.

The same reasoning gives 1/2 for a black card, since the other two suits, spades and clubs, are black.

Answer: A

The probability is 3/7.

Odds in favour of 3 : 4 mean 3 favourable outcomes for every 4 unfavourable ones, 7 in all. So the probability of the event is 3/(3 + 4) = 3/7.

4/7 is the probability of the event not happening, 3/4 is the odds written as a fraction, and 7/3 is greater than 1, which no probability can be.

In general, odds of a : b in favour give a probability of a/(a + b).

Answer: C

A binomial distribution has two outcomes, success or failure, on each trial.

The binomial distribution counts successes in a fixed number n of independent trials, each with only two outcomes and the same probability of success p.

The trials are independent, not dependent, and the count of successes is discrete, not continuous. The mean (np) need not equal the mode, so that is not a defining feature.

Answer: A

The probability is \(pq + (1-p)(1-q)\).

A computer is declared faulty in two ways:

• it is faulty (probability p) and the test is correct (probability q): \(pq\);
• it is good (probability 1 - p) and the test is wrong (probability 1 - q): \((1-p)(1-q)\).

Adding the two disjoint cases gives the answer. \(pq\) alone counts only the correctly detected faults, while \((1-q)p\) and \((1-p)q\) are the cases where a faulty machine passes or a good machine is correctly passed.

Answer: C

\(E(X^{3}) = \tfrac{1}{4}\).

For X uniform on (0, 1) the density is 1, so

\[ E(X^{3}) = \int_{0}^{1} x^{3}\,dx = \frac{1}{4} \]

1/2 is \(E(X)\) and 1/8 is \(E(X)^{3}\), the cube of the mean, which is not the mean of the cube. Zero would need X to be symmetric about 0.

In general, for X uniform on (0, 1), the mean of the k-th power is 1/(k + 1), so the second moment is 1/3 and the third 1/4.

Answer: D

The probability is \({}^{n}C_{k}\, p^{k}(1-p)^{n-k}\).

Any particular sequence of k successes and n - k failures has probability \(p^{k}(1-p)^{n-k}\), and there are \({}^{n}C_{k}\) ways of choosing which k of the n trials are the successes.

Without the \({}^{n}C_{k}\) factor you get the probability of one particular sequence only. \(p(1-p)^{k}\) and \({}^{n}C_{k}\, p^{n}(1-p)^{k}\) put the exponents in the wrong places.

Answer: A

The probability is about 0.68.

By Bayes' theorem, with G = girl and M = likes math:

\[ \begin{aligned} P(G \mid M) &= \frac{P(M \mid G)\,P(G)}{P(M)} \\ &= \frac{0.8 \times 0.6}{0.7} = \frac{0.48}{0.7} \approx 0.686 \end{aligned} \]

0.68 is the closest option. 0.48 would be the joint probability of a girl who likes math, which is not offered, and 0.25, 0.50 and 0.42 do not follow from the data.

Answer: C

The probability is 0.60.

Given that the student plays football, we look only at the 50 football players, of whom 30 also play cricket:

\[ P(C \mid F) = \frac{30}{50} = 0.60 \]

0.30 is the unconditional probability of playing both (30 out of 100), and 0.15 and 0.18 do not follow from the figures.

The general rule is P(C | F) = P(C and F) / P(F) = 0.30 / 0.50, which gives the same 0.60.

Answer: B

The variance is 52.083.

For a uniform distribution on (a, b) the variance is \((b-a)^{2}/12\):

\[ \frac{(30 - 5)^{2}}{12} = \frac{625}{12} \approx 52.083 \]

17.5 is the mean, \((5 + 30)/2\), not the variance, and 1/5 and 85.052 do not come from the formula.

For any uniform distribution the mean is the midpoint (a + b)/2 and the variance depends only on the width b - a, so shifting the interval does not change the variance.

Answer: D

The median is \(\log(2)/\lambda\).

For an exponential distribution with rate \(\lambda\), \(P(X \le x) = 1 - e^{-\lambda x}\). The median m makes this one half:

\[ \begin{aligned} 1 - e^{-\lambda m} &= \tfrac{1}{2} \\ m &= \frac{\ln 2}{\lambda} \end{aligned} \]

Here log means the natural logarithm. \(\log(2)/2\lambda\) and \(2\log(2)/\lambda\) are off by a factor of two, and \(\log(3)/\lambda\) would be the point with probability 2/3 below it.