Probability and Statistics Questions and Answers

Answer: B

The events \(X\) and \(Y\) are independent.

By symmetry, 1 precedes 2 in half of all permutations, so \(P(X) = 1/2\), and likewise \(P(Y) = 1/2\). The relative order of 1 and 2 and the relative order of 3 and 4 involve disjoint pairs of elements, and all four combinations of orders are equally likely, so

\[ P(X \cap Y) = \frac{1}{4} = P(X)\,P(Y) \]

They are not mutually exclusive, since both can happen (for example 1, 2, 3, 4, ...). Neither must occur: both fail together with probability 1/4. And X is not more likely than Y, since both have probability 1/2.

Answer: B

The probability is 1/3.

After two steps the urn holds 4 balls. Equal numbers means 2 red and 2 blue, which needs one red draw and one blue draw.

\[ \begin{aligned} P(R \text{ then } B) &= \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6} \\ P(B \text{ then } R) &= \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6} \end{aligned} \]

After the first draw the urn holds 3 balls, two of the drawn colour, so the other colour has probability 1/3. The total is 1/6 + 1/6 = 1/3.

1/2 treats the second draw as if the urn were unchanged, which ignores the reinforcement. The other outcomes, 3 red and 1 blue or the reverse, have probability 1/3 each, which is a useful check: 1/3 + 1/3 + 1/3 = 1.

Answer: C

The probability is \(\frac{365 \times 364}{366 \times 366}\).

Take the three people one at a time. The first can have any of the 366 days. The second must avoid that day, and the third must avoid both earlier days:

\[ \begin{aligned} P &= \frac{366}{366} \times \frac{365}{366} \times \frac{364}{366} \\ &= \frac{365 \times 364}{366 \times 366} \end{aligned} \]

The option \(1 - \frac{365 \times 364}{366 \times 366}\) is the complement: the chance that at least two of the three share a birthday. \(\frac{1}{366^{3}}\) is the chance that the three birthdays are three particular stated days, and \(\frac{1}{366 \times 365 \times 364}\) does not correspond to the event asked about; both are far too small.

Answer: C

The probability is \(\frac{1}{2}\).

Two fair coins give four equally likely outcomes: HH, HT, TH and TT. Exactly 2 heads is HH, and exactly 2 tails is TT; the two events cannot happen together, so their probabilities add:

\[ P = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \]

\(\frac{1}{4}\) is the probability of only one of the two events, for example just HH. \(\frac{1}{3}\) comes from wrongly treating HH, TT and 'one of each' as equally likely, and \(\frac{3}{8}\) belongs to a three-coin question.