Probability Questions and Answers

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.

Answer: C

The probability that they meet is 7/16.

Let their arrival times be x and y minutes after 1 PM, each uniform on 0 to 60. They meet when |x - y| ≤ 15. In the 60 by 60 square of possible arrivals, the region where they miss is two corner triangles, each with legs 60 - 15 = 45:

\[ \begin{aligned} P(\text{miss}) &= \frac{2 \times \tfrac{1}{2} \times 45^{2}}{60^{2}} = \frac{2025}{3600} = \frac{9}{16} \\ P(\text{meet}) &= 1 - \frac{9}{16} = \frac{7}{16} \end{aligned} \]

9/16 is the probability that they do not meet. 1/4 is 15/60, the waiting time as a fraction of the hour, which ignores that either may arrive first, and 1/16 is its square.