Mathematics 9709/62 — May/June 2011

Walkthrough

We first identify the distribution. The experiment is repeated in blocks of 20 throws, and we are counting how many times a 5 appears. Each throw gives either a 5 or not a 5, with the same probability each time, so the number of 5s in 20 throws is modelled by a binomial distribution $X \sim B(20, p)$.

The statement that the average number of 5s is 4.8 is a statement about the mean of this distribution. For a binomial distribution, $E(X) = np$, so:

$$20p = 4.8$$ $$p = 0.24$$

Therefore the probability of getting a 5 on one throw is 0.24, and the probability of not getting a 5 is 0.76.

We want the probability that the number of 5s is less than three, i.e. $X = 0$, $X = 1$ or $X = 2$. These are separate possible outcomes, so their probabilities are added. The binomial probability formula gives:

$$P(X = r) = {}^{20}C_r (0.24)^r (0.76)^{20-r}$$

Putting $r = 0, 1, 2$:

$$P(X < 3) = 0.76^{20} + {}^{20}C_1(0.24)(0.76)^{19} + {}^{20}C_2(0.24)^2(0.76)^{18}$$

The first term is the probability that no throw is a 5, the second that exactly one throw is a 5, and the third that exactly two throws are 5s. Evaluating:

$$P(X < 3) \approx 0.00413 + 0.02610 + 0.07831 = 0.1085$$

Rounded to three significant figures, this is $0.109$.

Key Takeaways

• The mean of a binomial distribution is $E(X) = np$; this can be used to recover the unknown probability $p$ from a given average.
• The probability of exactly $r$ successes in $n$ independent trials is ${}^{n}C_r p^r (1-p)^{n-r}$.
• When asked for a probability "less than" a value, list all acceptable integer outcomes and sum their probabilities.

Common Mistakes

• Mistaking the mean 4.8 for the probability $p$; the mean must be divided by $n = 20$ to obtain $p$.
• Forgetting the binomial coefficients, especially for $P(X=2)$, and writing $0.24^2$ without choosing which two throws are 5s.
• Using $P(X \le 3)$ or including $X = 3$ when the question says "less than three".
• Using a normal approximation when the exact binomial sum is expected; the normal approximation only appears as a special rule and would lose the method mark for the exact binomial sum.

Things to Be Careful About

• "Less than three" means $X = 0, 1, 2$; it does not include $X = 3$.
• Use $0.76$ for "not a 5", since $1 - 0.24 = 0.76$.
• Show the full unsimplified binomial expression before evaluating, as the mark scheme awards a mark for the sum of two or three binomial probabilities.
• Give the final answer to three significant figures, $0.109$, as required.