Probability Questions and Answers

Answer: B

The probability is \(\frac{1}{3}\).

The brother is selected with probability \(\frac{1}{6}\) (rejected \(\frac{5}{6}\)); the sister is rejected with probability \(\frac{3}{4}\), so selected with \(\frac{1}{4}\). Exactly one is selected if the brother is in and the sister out, or the sister in and the brother out:

\[ \begin{aligned} \tfrac{1}{6} \times \tfrac{3}{4} + \tfrac{5}{6} \times \tfrac{1}{4} &= \tfrac{3}{24} + \tfrac{5}{24} \\ &= \tfrac{8}{24} = \tfrac{1}{3} \end{aligned} \]

\(\frac{1}{4}\) is only the sister's chance, and \(\frac{5}{6}\) only the brother's chance of rejection.

Answer: D

The probability is \(\frac{3}{4}\).

The cards run from 12 to 71, so there are 71 - 12 + 1 = 60 cards. The primes among them are 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67 and 71, which is 15 numbers.

\[ \begin{aligned} P(\text{not prime}) &= \frac{60 - 15}{60} \\ &= \frac{45}{60} = \frac{3}{4} \end{aligned} \]

The options with denominator 59 miscount the cards (forgetting to include both ends), and \(\frac{1}{4}\) is the probability of drawing a prime.

Answer: A

The probability of getting a head is 3/5.

This is an experimental probability, the number of times the event happened divided by the number of trials. Counting the heads in H, T, H, T, T, H, H, T, H, H gives 6 heads in 10 tosses:

\[ P(\text{Head}) = \frac{6}{10} = \frac{3}{5} \]

2/5 is the probability of a tail (4 tails out of 10), the opposite event. 4/5 and 1/5 do not match either count.

Answer: A

The probability is \(\dfrac{12}{85}\).

The slips carry the whole numbers from 8 to 92 inclusive:

\[ 92 - 8 + 1 = 85 \text{ slips} \]

The multiples of 7 in that range run from \(14 = 7 \times 2\) to \(91 = 7 \times 13\), which is \(13 - 2 + 1 = 12\) numbers. So

\[ P = \frac{12}{85} \]

\(\frac{13}{85}\) wrongly counts 7, which is below 8. \(\frac{13}{84}\) also miscounts the slips as \(92 - 8\), leaving out one end. \(\frac{1}{7}\) assumes exactly one number in seven, which is not true for this range.

Answer: D

The equally likely situation is captains of two teams tossing a coin to decide which team will bat first.

A fair coin has two outcomes, head and tail, each with probability 1/2, so each captain has the same chance of winning the toss.

The others are not equally likely. Whether a train is on time or late depends on many conditions, and the two outcomes need not have equal chances. A waitlisted ticket's chance of confirmation depends on its position and cancellations. A student who prepared 3 chapters out of 5 has a 3/5 chance of getting a question from them, not 1/2.

Answer: C

The probability is \(\frac{3}{26}\).

A deck of 52 cards has four 10s and two red kings (hearts and diamonds). These cards do not overlap, so the favourable outcomes simply add.

\[ P = \frac{4 + 2}{52} = \frac{6}{52} = \frac{3}{26} \]

\(\frac{1}{13}\) counts only the 10s, \(\frac{1}{26}\) counts only the red kings, and \(\frac{2}{13}\) counts all four kings instead of the two red ones.

Answer: D

The probability that the ball is not red is \(\frac{1}{2}\).

There are 3 + 2 + 5 = 10 balls in all, of which 5 are red, so 5 are not red (3 white and 2 blue):

\[ P(\text{not red}) = \frac{5}{10} = \frac{1}{2} \]

Equivalently, \(1 - P(\text{red}) = 1 - \frac{5}{10}\). \(\frac{3}{10}\) is the probability of white alone, \(\frac{1}{5}\) is the probability of blue alone, and \(\frac{4}{5}\) does not correspond to any event here.

Answer: A

The probability that B hits is about 0.14.

Let B hit with probability p, so A hits with probability 3p. The shots are independent, and the convict escapes injury only if both miss:

\[ \begin{aligned} (1 - 3p)(1 - p) &= 0.5 \\ 3p^{2} - 4p + 0.5 &= 0 \\ p &= \frac{4 - \sqrt{10}}{6} \approx 0.14 \end{aligned} \]

The other root, about 1.19, is not a probability. Check: A hits with 0.42, and \(0.58 \times 0.86 \approx 0.50\). The value 0.33 would make A's probability 0.99, and 0.22 or 0.40 give a non-injury probability far from 0.5.

Answer: A

The probability is about 0.2508.

Each attempt succeeds with probability 0.6, independently, so the number of successes in 10 attempts is binomial:

\[ \begin{aligned} P(6) &= \binom{10}{6}(0.6)^{6}(0.4)^{4} \\ &= 210 \times 0.046656 \times 0.0256 \\ &\approx 0.2508 \end{aligned} \]

0.6000 is the success rate per attempt, not the chance of exactly 6 in 10. The other values come from errors in the binomial coefficient or the powers; even the most likely count of a binomial has probability well below 0.6.

Answer: B

The most likely statement is One H and one T will occur.

Tosses of an unbiased coin are independent, so the earlier trials and the first four tosses of the fourth trial tell us nothing about the last two. Those two tosses have four equally likely outcomes: HH, HT, TH and TT, each with probability 1/4.

'One H and one T' is satisfied by both HT and TH, so its probability is 1/2. 'Two T' (TT only) and 'Two H' (HH only) each have 1/4. 'One H followed by one T' fixes the order, so it covers HT alone and also has 1/4.