一个骰子是有偏差的，且已知掷出六点的概率 $p$ 小于 $\frac{1}{6}$。一项实验记录了投掷该骰子 25 次中出现六点的次数。在大量实验中，出现六点次数的标准差为 1.5。

1. Set up the equation for the variance of the binomial distribution.

Let $X$ be the number of sixes in 25 throws. We have $X \sim \mathrm{B}(25, p)$.
The standard deviation is $1.5$, so the variance is $1.5^2 = 2.25$.
$n p (1 - p) = \mathrm{Var}(X)$
$25 p (1 - p) = 2.25$

2. Solve the quadratic equation to show $p = \frac{1}{10}$.

$p - p^2 = \frac{2.25}{25} = 0.09$
$p^2 - p + 0.09 = 0$
$(p - 0.1)(p - 0.9) = 0$
So $p = 0.1$ or $p = 0.9$.
Since we are given that $p < \frac{1}{6}$, we reject $p = 0.9$.
Thus, $p = 0.1 = \frac{1}{10}$ (shown).

3. Formulate the required probability expression.

We want to find the probability that at least 6 but fewer than 10 sixes are recorded. This corresponds to $\mathrm{P}(6 \le X < 10)$.
Since $X$ is discrete, this is equivalent to $\mathrm{P}(6 \le X \le 9)$.

4. Evaluate the probability using a graphing calculator.

Using a graphing calculator,
$\mathrm{P}(6 \le X \le 9) = \mathrm{P}(X \le 9) - \mathrm{P}(X \le 5)$
$\mathrm{P}(6 \le X \le 9) \approx 0.99992 - 0.96660$
$\mathrm{P}(6 \le X \le 9) = 0.0333 \text{ (to 3 s.f.)}$

1. Define the new binomial random variable and state its distribution.

From the previous part, the probability of throwing a six is $p = \frac{1}{10} = 0.1$.
Let $Y$ be the number of sixes obtained in 40 throws.
$Y \sim \text{B}(40, 0.1)$

2. Identify the most likely number of sixes by comparing probabilities around the mean.

The expected number of sixes is $\text{E}(Y) = np = 40 \times 0.1 = 4$.
We use a graphing calculator to compare the probabilities $P(Y = y)$ for integers around the mean:

• $P(Y = 3) \approx 0.2003$
• $P(Y = 4) \approx 0.2059$
• $P(Y = 5) \approx 0.1647$

Since $P(Y = 4)$ is the highest, the most likely number of sixes obtained is 4.