Solution — 2022 Question 11 Probability| Binomial Distribution

(a)

Let $S$ represent a successfully attempt and $F$ represent a unsuccessfully attempt.

(b)

$\begin{matrix} P (completes challenge) \\ = 0.4 + 0.6 (0.2) + 0.6 (0.8) (0.2) \\ = 0.4 + 0.12 + 0.096 \\ = 0.616 ∎ \end{matrix}$

(c)

$\begin{matrix} P (completes | not successful at first) \\ = \frac{P (completes \cap not successful at first)}{P (not successful at first)} \\ = \frac{0.6 (0.2) + 0.6 (0.8) (0.2)}{0.6} \\ = \frac{0.12 + 0.096}{0.6} \\ = \frac{0.216}{0.6} \\ = 0.36 ∎ \end{matrix}$

(d)

Let $X$ represent the r.v. of the number of challenges completed successfully out of $10.$

$ ParseError: Can't use function '$' in math mode at position 1: $̲ X \sim \mathrm{B} \left( 10, 0.616 \right)

\begin{align*} \mathrm{P}\left(X = 6\right) &= 0.249\,48 \\ &= 0.249\; \QED \text{ (to 3 s.f.)} \end{align*}$$ ParseError: {align*} can be used only in display mode.

[probability of X = 6]

(e)

$\begin{matrix} P (X \geq 4) & = 1 - P (X \leq 3) \\ = 1 - 0.043 794 \\ = 0.956 21 \\ = 0.956 ∎ (to 3 s.f.) \end{matrix}$

[probability of X \geq 4]

(f)

Let $Y$ represent the number of students who completed the challenge out of $12.$

$ ParseError: Can't use function '$' in math mode at position 1: $̲ Y \sim \mathrm{B} \left( 12, 0.956,21 \right)

\begin{align*} \mathrm{P}\left(Y < 9\right) &= \mathrm{P}\left(Y \leq 8\right) \\ &= 0.001\,371\,0 \\ &= 0.00137\; \QED \text{ (to 3 s.f.)} \end{align*}$$ ParseError: {align*} can be used only in display mode.

[probability of Y < 9]