Probability Questions and Answers

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.

Answer: A

No more than 700 kits are needed.

Step (ii) tests each of the \(1000/5 = 200\) pooled sets once: 200 kits. Step (iv) then tests the 5 members of every set whose pool is positive. The worst case is when the 100 positive people are spread one per set, making 100 positive pools:

\[ 200 + 100 \times 5 = 700 \]

Any grouping that puts two or more positives in the same set gives fewer positive pools and so needs fewer kits, which is why 700 is an upper bound 'irrespective of how they are grouped'. 600 assumes only 80 positive pools, 800 and 1000 overcount, and 1000 is what testing everyone individually would cost.

Answer: A

The probability is 3/8.

Three children give 2³ = 8 equally likely sequences. Exactly two girls and one boy occur in GGB, GBG and BGG, so

\[ P = \binom{3}{2}\left(\tfrac{1}{2}\right)^{3} = \tfrac{3}{8} \]

1/8 is the probability of one particular order, which forgets that the boy can be first, second or third. 1/2 and 1/4 do not follow from the count.

Answer: B

The true statement is Y and Z are dependent.

There are 8 equally likely outcomes. \(X = \{HHH, HHT\}\), \(Y\) = tail on the third toss (4 outcomes), \(Z\) = exactly two tails = \(\{HTT, THT, TTH\}\).

\[ \begin{aligned} P(Y)P(Z) &= \tfrac{1}{2} \cdot \tfrac{3}{8} = \tfrac{3}{16} \\ P(Y \cap Z) &= P(\{HTT, THT\}) = \tfrac{2}{8} = \tfrac{4}{16} \end{aligned} \]

These differ, so Y and Z are dependent, which also rules out 'Y and Z are independent'. For X and Y, \(P(X \cap Y) = P(HHT) = \frac{1}{8} = \frac{1}{4} \cdot \frac{1}{2}\), so they are independent and 'X and Y are not independent' is false. X and Z cannot both happen, while each has positive probability, so they are dependent.

Answer: C

The probability is \(\frac{175}{6^{5}}\).

The experiment ends at trial 5 exactly when trials 4 and 5 are both sixes and the experiment has not already ended. That requires trial 3 not to be a six (otherwise trials 3 and 4 would end it at trial 4), and trials 1 and 2 not to be both sixes (which would end it at trial 2). Trials 2 and 3 cannot both be six because trial 3 is not.

\[ \begin{aligned} P &= \tfrac{35}{36} \cdot \tfrac{5}{6} \cdot \tfrac{1}{6} \cdot \tfrac{1}{6} \\ &= \frac{35 \cdot 5}{6^{5}} = \frac{175}{6^{5}} \end{aligned} \]

125 comes from requiring trials 1 to 3 all to be non-sixes, which wrongly excludes sequences such as 6, 1, 2, 6, 6; 150 and 200 do not come from a correct count.

Answer: B

The probability is 3/8.

After three black balls are added there are seven balls, so five are black exactly when two of the original four are black. Each original ball is black with probability 1/2, independently, so the number of black ones among them is binomial:

\[ \begin{aligned} P &= \binom{4}{2}\left(\frac{1}{2}\right)^{4} \\ &= \frac{6}{16} = \frac{3}{8} \end{aligned} \]

5/7 and 2/7 come from treating 'five of seven black' as a ratio of balls rather than a probability over the unknown colours. 1/2 is the chance for a single original ball, not for exactly two of the four.

Answer: B

The probability is 1/55.

There are 12 balls, 4 of them red, and the draws are without replacement:

\[ \begin{aligned} P &= \frac{4}{12} \times \frac{3}{11} \times \frac{2}{10} \\ &= \frac{24}{1320} = \frac{1}{55} \end{aligned} \]

The same answer comes from combinations: \(\binom{4}{3}/\binom{12}{3} = 4/220\). 1/27 is \((1/3)^{3}\), the answer with replacement. 1/36 and 1/72 come from other slips in the counting.

Answer: C

The probability is about 0.33%.

"At least four" means exactly four ones or exactly five, using the binomial distribution with p = 1/6:

\[ \begin{aligned} P &= \binom{5}{4}\left(\frac{1}{6}\right)^{4}\frac{5}{6} + \left(\frac{1}{6}\right)^{5} \\ &= \frac{25 + 1}{7776} = 0.00334 \end{aligned} \]

As a percentage that is 0.334%, which rounds to 0.33. 3.33 and 33.3 are ten and a hundred times too large, the decimal slips of not converting correctly to a percentage, and 0.0033 is the probability itself left as a fraction rather than a percentage.

Answer: D

The probability is 1/2.

Drawing one after another without replacement gives the same chances as drawing three together, so count combinations:

\[ \begin{aligned} P &= \frac{\binom{4}{1}\binom{6}{2}}{\binom{10}{3}} = \frac{4 \times 15}{120} \\ &= \frac{60}{120} = \frac{1}{2} \end{aligned} \]

The same answer comes from ordered draws: the chance of red, black, black in that order is \(\frac{4}{10}\cdot\frac{6}{9}\cdot\frac{5}{8} = \frac{1}{6}\), and the red ball can come in any of 3 positions, giving 1/2. Forgetting those 3 positions gives 1/6, which is not offered; 1/20, 1/12 and 3/10 do not match the count.

Answer: B

The probability is about 0.334.

Use Bayes' theorem. The probability that a shock absorber is reliable is:

\[ \begin{aligned} P(R) &= 0.6(0.96)+0.4(0.72) \\ &= 0.576+0.288 = 0.864 \end{aligned} \]

\[ P(Y \mid R) = \frac{0.288}{0.864} = \frac{1}{3} = 0.333 \]

The listed 0.334 is the nearest option to 1/3. 0.288 is the joint probability P(Y and R), not the conditional one; 0.667 is P(X | R); 0.720 is P(R | Y), the reverse conditional.