Solution — 2024 Question 10 Probability| Binomial Distribution

(a)

tree diagram

(b)

\begin{matrix} P (hits exactly once) \\ = 0.7 (0.2) (0.6) + 0.3 (0.4) (0.2) + 0.3 (0.6) (0.4) \\ = 0.18 ∎ \end{matrix}

(c)

\begin{matrix} P (at least 1 token | missed 2nd) \\ = \frac{P (at least 1 token \cap missed 2nd)}{P (missed 2nd)} \\ = \frac{0.7 (0.2) (0.6) + 0.7 (0.2) (0.4) + 0.3 (0.6) (0.4)}{0.7 (0.2) + 0.3 (0.6)} \\ = 0.6625 \\ = \frac{53}{80} ∎ \end{matrix}

(d)

\begin{matrix} P (6 tokens) \\ = 0.7 (0.8) (0.8) \\ = 0.448 \end{matrix}

Let $Y$ represent the number of friends who get $6$ tokens (hit target 3 times) out of $4$ friends.

$Y \sim B (4, 0.448)$

\begin{matrix} P (Y \geq 2) & = 1 - P (Y \leq 1) \\ = 1 - 0.394 25 \\ = 0.605 75 \\ = 0.606 ∎ (to 3 s.f.) \end{matrix}

probability of Y \geq 2