Statistics and Probability Questions and Answers

Answer: C

The probability is \(\frac{1}{240}\).

STATISTICS has S three times, T three times, I twice, and A and C once each.

\[ \begin{aligned} \text{total} &= \frac{10!}{3!\,3!\,2!} \\ &= \frac{3628800}{72} = 50400 \end{aligned} \]

If the arrangement starts with C, A, T, the remaining 7 letters are S, S, S, T, T, I, I:

\[ \begin{aligned} \text{favourable} &= \frac{7!}{3!\,2!\,2!} = 210 \\ P &= \frac{210}{50400} = \frac{1}{240} \end{aligned} \]

Dividing 50400 by 210 gives exactly 240. Counting the remaining 7 letters as if all were different (7! = 5040) would give \(\frac{1}{10}\), far from every option, which is why the repeated S, T and I must be divided out.

Answer: B

The mean deviation is 4.5.

The mean is \(\frac{5 + 6 + 14 + 15}{4} = 10\). The absolute deviations from 10 are 5, 4, 4, 5:

\[ \text{M.D.} = \frac{5 + 4 + 4 + 5}{4} = \frac{18}{4} = 4.5 \]

Dividing by 5 instead of 4 gives 3.6, and the other options do not come from these deviations.

Answer: A

The variance is \(\frac{50}{3}\).

The mean is 70, and the squared deviations are \(25, 0, 25\):

\[ \sigma^{2} = \frac{25 + 0 + 25}{3} = \frac{50}{3} \]

\(\frac{50}{2}\) is the sample variance (dividing by \(n - 1\)), which is not the definition used here. 70 is the mean, and \(\frac{50}{6}\) halves the result.

Answer: E

The statement that is not true is \(\{1, 3, 2\}\) and \(\{2, 4, 6\}\) are mutually exclusive events.

Two events are mutually exclusive when they cannot happen together, that is, when they have no outcome in common. Both of these sets contain 2, so a roll of 2 makes both happen.

The others are true. \(\{1, 3\}\) and \(\{2, 4, 6\}\) share nothing. \(\{1, 5\}\), \(\{2, 4\}\) and \(\{3, 6\}\) are disjoint and together cover all six outcomes, so they are mutually exclusive and exhaustive. \(\{1, 2, 3, 4, 5, 6\}\) is the whole sample space, the sure event. Single-outcome sets such as \(\{1\}\) are elementary events.

Answer: E

The impossible assignment is \(P(A) = \frac{77}{190}, P(B) = \frac{47}{190}, P(C) = \frac{67}{190}\).

If \(A\), \(B\), \(C\) are mutually exclusive and are all the possible events, they are exhaustive, so their probabilities must add up to exactly 1.

\[ \frac{77 + 47 + 67}{190} = \frac{191}{190} \gt 1 \]

The other sets sum to exactly 1: \(\frac{19}{19}\), \(\frac{95}{95}\), \(\frac{190}{190}\) and \(\frac{95}{95}\), and every individual value lies between 0 and 1, so they are permissible.

Answer: D

\(\sum f_{i}\) is 40.

The mean deviation about the mean for a frequency distribution is

\[ \text{M.D.} = \frac{\sum f_{i}|x_{i} - \bar{x}|}{\sum f_{i}} \]

So \(10 = \frac{400}{\sum f_{i}}\), giving \(\sum f_{i} = 40\). The other values would give mean deviations of about 7.3, 8, 11.1 and 16.

Answer: C

The standard deviation is about 8.136.

The sum runs over 10 values, so there are 10 observations. The variance is the mean of the squared deviations:

\[ \begin{aligned} \sigma^{2} &= \frac{662}{10} = 66.2 \\ \sigma &= \sqrt{66.2} \approx 8.136 \end{aligned} \]

Check: \(8.136^{2} \approx 66.19\). The other values square to roughly 41.6, 83.3, 83.6 and 50.6, none of which is 66.2.

Answer: A

\(P(B)\) is \(\dfrac{1}{5}\).

Both conditional probabilities involve the same \(P(A \cap B)\):

\[ \begin{aligned} P(A \cap B) &= P(B/A)P(A) = \tfrac{2}{5} \times \tfrac{1}{8} = \tfrac{1}{20} \\ P(B) &= \frac{P(A \cap B)}{P(A/B)} = \frac{1/20}{1/4} = \frac{1}{5} \end{aligned} \]

\(\frac{2}{5}\) is \(P(B/A)\) itself, a common slip when the two conditional probabilities are confused.

Answer: E

The probability is 0.46.

The expression \(\frac{(x - 15)(x - 70)}{x - 30}\) changes sign at 15, 30 and 70. Testing each interval:

• \(x \lt 15\): one positive factor out of three, so negative.
• \(15 \leq x \lt 30\): non-negative (zero at 15).
• \(30 \lt x \lt 70\): negative.
• \(x \geq 70\): non-negative (zero at 70).

\(x = 30\) is excluded because the denominator vanishes. The favourable numbers are 15 to 29 (15 numbers) and 70 to 100 (31 numbers), 46 in all, so the probability is \(\frac{46}{100} = 0.46\). Leaving out the end points 15 or 70 gives 0.45 or less, and including 30 gives 0.47.

Answer: A

The probability is \(\frac{1}{9}\).

Given that the first throw is 5, the remaining two throws must add to 9. The pairs that do so are (3, 6), (4, 5), (5, 4) and (6, 3): 4 of the 36 equally likely outcomes.

\[ P = \frac{4}{36} = \frac{1}{9} \]

Because the first throw is given, only the last two throws are random, so the sample space has 36 outcomes, not 216. The other fractions do not match 4 favourable outcomes out of 36.