Solution (a)

Step 1 (Formula / Governing Law & Given Data): [1 Mark]

Let $p$ be the probability of hitting the target and $q$ be the probability of missing the target. Given $p = 3q$. Since $p + q = 1$, we have $3q + q = 1 \implies 4q = 1$.

• Probability of hitting ($p$) = $\frac{3}{4}$
• Probability of missing ($q$) = $\frac{1}{4}$

Step 2 (Step-by-Step Mathematical Working): [1.5 Marks]

The sniper fires two independent shots. Let $H$ denote a hit and $M$ denote a miss.

(i) Probability that the target is hit (at least once):

$$P(\text{at least one hit}) = 1 - P(\text{no hits}) = 1 - (q \times q) = 1 - \left(\frac{1}{4}\right)^2 = 1 - \frac{1}{16} = \frac{15}{16}$$

(ii) Probability that at least one shot misses the target:

$$P(\text{at least one miss}) = 1 - P(\text{no misses}) = 1 - P(\text{both hits}) = 1 - (p \times p) = 1 - \left(\frac{3}{4}\right)^2 = 1 - \frac{9}{16} = \frac{7}{16}$$

Step 3 (Final Concluded Answer): [0.5 Mark]

(i) Probability target is hit = $\frac{15}{16}$

(ii) Probability at least one shot misses = $\frac{7}{16}$

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Solution (b)

Step 1 (Formula / Governing Law & Given Data): [1 Mark]

The sample space $S$ for 3 people (M, F, S) is: $\{MFS, MSF, FMS, FSM, SMF, SFM\}$. Total outcomes $n(S) = 6$.

Event $F$ (Father in middle): $\{MFS, SFM\}$. $n(F) = 2$.

Event $E$ (Son on one end): $\{SMF, SFM, FMS, MSF\}$.

Formula: $P(E/F) = \frac{P(E \cap F)}{P(F)}$

Step 2 (Step-by-Step Mathematical Working): [1.5 Marks]

Find $E \cap F$ (Son on one end AND Father in middle):

From the set $F = \{MFS, SFM\}$, we check which satisfy $E$:

• $MFS$: Son is at the end. (Yes)
• $SFM$: Son is at the end. (Yes)

So, $E \cap F = \{MFS, SFM\}$, $n(E \cap F) = 2$.

$$P(E \cap F) = \frac{2}{6}, \quad P(F) = \frac{2}{6}$$ $$P(E/F) = \frac{2/6}{2/6} = 1$$

Step 3 (Final Concluded Answer): [0.5 Mark]