Probability Questions and Answers

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.

Answer: D

The probability is 7/64.

Each disk stops on one of 8 equally likely numbers, independently, so there are \(8 \times 8 = 64\) equally likely pairs. The pairs adding to 8 are

(1, 7), (2, 6), (3, 5), (4, 4), (5, 3), (6, 2), (7, 1)

that is, 7 pairs; (8, 0) is impossible because no sector is numbered 0. The probability is therefore 7/64.

3/32, which is 6/64, leaves out the pair (4, 4); 1/16, which is 4/64, counts each unordered pair such as {3, 5} only once; and 5/64 also undercounts the seven ordered pairs.

Answer: C

The probability is 2/7.

A leap year has 366 days, which is 52 complete weeks plus 2 extra days. The 52 weeks give exactly 52 Sundays, so a 53rd Sunday occurs only if one of the two extra days is a Sunday.

The two extra days are consecutive days, and there are 7 equally likely pairs: Sunday-Monday, Monday-Tuesday, Tuesday-Wednesday, Wednesday-Thursday, Thursday-Friday, Friday-Saturday and Saturday-Sunday. Two of them contain a Sunday, so the probability is 2/7.

1/7 is the answer for an ordinary year of 365 days, which has only one extra day. 1/4 and 4/7 have no basis in the count.

Answer: A

The probability is \(\frac{75}{216}\).

In three rolls, the single 6 can come on any one of the 3 rolls, and the other two rolls must each show something other than 6. By the binomial rule,

\[ \begin{aligned} P &= 3 \times \frac{1}{6} \times \left(\frac{5}{6}\right)^{2} \\ &= 3 \times \frac{25}{216} = \frac{75}{216} \end{aligned} \]

\(\frac{25}{216}\) is the probability of one particular order, such as a 6 on the first roll only, and forgets the factor 3. \(\frac{1}{6}\) is the chance of a 6 on a single roll, and \(\frac{1}{18}\) does not correspond to the event asked for.

Answer: C

The probability is \(\frac{7}{18}\).

There are 36 equally likely ordered pairs (first, second). Count the pairs where the second number is a multiple of the first:

• first 1: 1, 2, 3, 4, 5, 6 (6 pairs)
• first 2: 2, 4, 6 (3 pairs)
• first 3: 3, 6 (2 pairs)
• first 4, 5 or 6: only itself (1 pair each, 3 in all)

\[ P = \frac{6+3+2+3}{36} = \frac{14}{36} = \frac{7}{18} \]

1/6 counts only the six pairs where both rolls match, and 5/18 also undercounts; 5/6 is far too large, since a large first roll such as 4, 5 or 6 allows only one second roll.

Answer: B

The probability is \(\frac{5}{324}\).

Six dice have \(6^{6} = 46656\) equally likely outcomes. All six faces appear exactly once in \(6! = 720\) of them, one for each ordering of the numbers 1 to 6 over the dice.

\[ P = \frac{720}{46656} = \frac{5}{324} \approx 0.0154 \]

\(\frac{1}{324}\) undercounts the orderings by a factor of 5, and \(\frac{7}{324}\) and \(\frac{11}{324}\) do not reduce from \(6!/6^{6}\).

Answer: A

The probability is 1/4.

Each day is cloudy with probability 1/2, independently, so the number of cloudy days in four is binomial. Exactly three cloudy days can happen in C(4, 3) = 4 ways (the sunny day can be any of the four):

\[ \begin{aligned} P &= \binom{4}{3}\left(\tfrac{1}{2}\right)^{3}\left(\tfrac{1}{2}\right)^{1} \\ &= 4 \times \tfrac{1}{16} = \tfrac{1}{4} \end{aligned} \]

3/8 is the probability of exactly two cloudy days, C(4, 2)/16 = 6/16, and is the usual slip. 3/4 and 2/3 do not come from any count of the 16 equally likely weather sequences.

Answer: C

The probability is \(\frac{15}{36}\).

Two rolls give 36 equally likely ordered outcomes, with sums from 2 to 12. The prime sums are 2, 3, 5, 7 and 11. Counting the ordered pairs for each:

• sum 2: (1,1), 1 way;
• sum 3: (1,2), (2,1), 2 ways;
• sum 5: 4 ways;
• sum 7: 6 ways;
• sum 11: (5,6), (6,5), 2 ways.

The total is 1 + 2 + 4 + 6 + 2 = 15, so the probability is 15/36. 3/36 and 13/36 leave out some of the prime sums, and 19/36 includes outcomes whose sum is not prime.

Answer: C

The most likely outcome is Five green faces and two red faces.

Each roll shows green with probability 2/3 and red with probability 1/3, so the number of greens in 7 rolls is binomial. The chance of \(k\) greens is proportional to \(\binom{7}{k} 2^{k}\):

Five greens gives the largest value. The usual shortcut is that the most likely count of a binomial is the whole-number part of \((n + 1)p = 8 \times \frac{2}{3} = 5.33\), which is 5. Six greens is close but has fewer arrangements, and three or four greens underweight the more likely colour.

Answer: D

The probability is \(\frac{23}{50}\).

Condition on the colour of the first ball. If it is orange (probability \(\frac{2}{5}\)), it is replaced, so the box is unchanged and the second draw is orange with probability \(\frac{2}{5}\). If it is green (probability \(\frac{3}{5}\)), it is not replaced, leaving 2 green and 2 orange, so the second draw is orange with probability \(\frac{1}{2}\).

\[ \begin{aligned} P &= \tfrac{2}{5} \cdot \tfrac{2}{5} + \tfrac{3}{5} \cdot \tfrac{1}{2} \\ &= \tfrac{8}{50} + \tfrac{15}{50} = \tfrac{23}{50} \end{aligned} \]

\(\frac{8}{25}\) is the orange-then-orange term alone counted twice, and \(\frac{1}{2}\) is only the green branch's conditional probability. \(\frac{19}{50}\) does not correspond to either case.