Statistics and Probability Questions and Answers

Answer: A

The probability is \(\frac{2}{105}\).

First draw, 3 white from 6 white among 10 balls; second draw, 3 red from the 4 red among the 7 that remain:

\[ \begin{aligned} P &= \frac{{}^{6}C_{3}}{{}^{10}C_{3}} \times \frac{{}^{4}C_{3}}{{}^{7}C_{3}} \\ &= \frac{20}{120} \times \frac{4}{35} = \frac{4}{210} = \frac{2}{105} \end{aligned} \]

Forgetting that the first draw leaves only 7 balls (using \({}^{10}C_{3}\) twice) gives \(\frac{1}{6} \times \frac{1}{30} = \frac{1}{180}\); \(\frac{4}{105}\) doubles the answer.

Answer: C

\(P(A) = \) 0.49.

By the theorem of total probability, splitting on \(B\) and its complement \(B'\) with \(P(B') = 0.3\):

\[ \begin{aligned} P(A) &= P(A|B)P(B) + P(A|B')P(B') \\ &= 0.4(0.7) + 0.7(0.3) \\ &= 0.28 + 0.21 = 0.49 \end{aligned} \]

Using 0.7 for \(P(B')\) as well gives 0.77; averaging the two conditional probabilities gives 0.55; 0.44 and 0.54 are similar slips.

Answer: D

The standard deviation is \(\frac{\sqrt{66}}{2}\).

The mean is \(\frac{-3 + 0 + 3 + 8}{4} = 2\). The squared deviations are

\[ \begin{aligned} (-5)^{2} + (-2)^{2} + 1^{2} + 6^{2} &= 66 \\ \sigma^{2} = \tfrac{66}{4},\quad \sigma &= \tfrac{\sqrt{66}}{2} \end{aligned} \]

Check with the other formula:

\[ \begin{aligned} \sigma^{2} &= \overline{x^{2}} - \bar{x}^{2} = \frac{9 + 0 + 9 + 64}{4} - 4 \\ &= \frac{82 - 16}{4} = \frac{66}{4} \end{aligned} \]

The neighbouring surds 60, 62, 65 and 67 come from an arithmetic slip in one square.

Answer: C

The ratio is \(\frac{216}{125}\).

The first 5 on toss \(k\) means \(k - 1\) failures, each with probability \(\frac{5}{6}\), then a success with probability \(\frac{1}{6}\): \(P(X = k) = \left(\frac{5}{6}\right)^{k - 1}\frac{1}{6}\).

\[ \begin{aligned} \frac{P(X = 2)}{P(X = 5)} &= \frac{(5/6)^{1}}{(5/6)^{4}} \\ &= \left(\frac{6}{5}\right)^{3} = \frac{216}{125} \end{aligned} \]

\(\frac{125}{216}\) is the reciprocal (the ratio taken the other way up), and \(\frac{36}{25}\) uses a difference of 2 failures instead of 3.

Answer: A

\(P(A) =\) \(\frac{7}{32}\).

\(X \times Y\) has \(2 \times 4 = 8\) elements, so it has \(2^{8} = 256\) subsets, all equally likely. Those with exactly three elements number \({}^{8}C_{3} = 56\):

\[ P(A) = \frac{56}{256} = \frac{7}{32} \]

\(\frac{7}{64}\) and \(\frac{7}{128}\) use \(2^{9}\) or \(2^{10}\) subsets, and the options with 5 in the numerator miscount \({}^{8}C_{3}\).

Answer: E

The probability is \(\frac{1}{7}\).

There are 35 balls. The first draw is white with probability \(\frac{10}{35}\); then 34 balls remain, 17 of them black:

\[ \begin{aligned} P &= \frac{10}{35} \times \frac{17}{34} \\ &= \frac{2}{7} \times \frac{1}{2} \\ &= \frac{1}{7} \end{aligned} \]

\(\frac{2}{7}\) is the first draw alone, \(\frac{17}{175}\) treats the draws as with replacement (\(\frac{10}{35}\cdot\frac{17}{35}\)), and \(\frac{2}{35}\) and \(\frac{5}{34}\) do not follow.

Answer: B

The probability is \(\frac{1}{4}\).

A product is odd only when both factors are odd. \(S\) has four odd numbers (5, 11, 17, 23) out of eight, so each draw is odd with probability \(\frac{1}{2}\), independently because the draws are with replacement:

\[ \begin{aligned} P(mn \text{ odd}) &= \frac{1}{2} \times \frac{1}{2} \\ &= \frac{1}{4} \end{aligned} \]

\(\frac{3}{4}\) is the probability that \(mn\) is even, \(\frac{1}{2}\) counts only one draw, and \(\frac{3}{7}\) treats the draws as without replacement.

Answer: D

\(n =\) 25.

Raising one mark from 50 to 75 adds 25 to the total, and the mean rises by \(66 - 65 = 1\). Since the total equals \(n\) times the mean:

\[ \begin{aligned} 66n &= 65n + 25 \\ n &= 25 \end{aligned} \]

Each of the other values would change the mean by \(\frac{25}{n}\), giving 2.5, 1.67, 1.25 or 0.83 instead of 1.

The general shortcut: a change \(\Delta\) in one mark changes the mean by \(\frac{\Delta}{n}\), so \(n = \frac{25}{1}\).

Answer: D

The mean deviation is 12.

The mean is 420. The absolute deviations from it are 20, 10, 0, 10, 20.

\[ \begin{aligned} \text{M.D.} &= \frac{20 + 10 + 0 + 10 + 20}{5} \\ &= \frac{60}{5} = 12 \end{aligned} \]

10 and 20 are single deviations, not their mean, and 14 or 16 would need a different sum. (The standard deviation of this data is \(\sqrt{200} \approx 14.1\), which is a different measure.)

Answer: D

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

A fair die has six equally likely faces, of which three, 1, 3 and 5, are odd.

\[ P(B) = \frac{3}{6} = \frac{1}{2} \]

\(\frac{1}{6}\) is the probability of one particular face, \(\frac{1}{3}\) of two faces, and \(\frac{1}{4}\) and \(\frac{1}{5}\) do not correspond to any count of the six faces.