Q.Three persons, $A$, $B$ and $C$, fire at a target in turn, starting with $A$. Their probabilities of hitting the target are $0.4$, $0.3$ and $0.2$ respectively. The probability of two hits is
(A) $0.024$
(B) $0.188$
(C) $0.336$
(D) $0.452$

Concept: Conditional Probability — each shot is independent, and the order of firing matters only for who shoots when.

Step 1 – List all ways to get exactly two hits

The three persons fire in order A, B, C. Two hits means one misses. The three cases:

• A and B hit, C misses
• A and C hit, B misses
• B and C hit, A misses

Step 2 – Compute each probability

Given $P(A)=0.4$, $P(B)=0.3$, $P(C)=0.2$, and misses are complements:

• Case 1: $0.4 \times 0.3 \times (1-0.2) = 0.4 \times 0.3 \times 0.8 = 0.096$
• Case 2: $0.4 \times (1-0.3) \times 0.2 = 0.4 \times 0.7 \times 0.2 = 0.056$ …

Why conditional probability isn’t needed here

A common trap is to think “they fire in turn, so later shooters depend on earlier ones.” But the problem states each person fires exactly once, and their hits are independent events. The order of firing doesn’t affect the probability of a given combination — it only determines who shoots when. So we simply treat A, B, and C as three independent trials with given success probabilities.

The phrase “in turn” just tells us the sequence, not that later shooters only fire if earlier ones miss. Each fires regardless. So the probability of two hits is the sum over all ways to choose which two hit and which one misses.

Step-by-step

1. List the three possible cases for exactly two hits

   • A and B hit, C misses
   • A and C hit, B misses
   • B and C hit, A misses

   No other combination gives exactly two hits.

2. Compute each case using independence

   For independent events, $P(\text{A hits and B hits and C misses}) = P(A) \cdot P(B) \cdot P(C^c)$.

   Similarly for the others.

   • Case 1: A hits (0.4), B hits (0.3), C misses ($1-0.2 = 0.8$)

• Case 2: A hits (0.4), B misses ($1-0.3 = 0.7$), C hits (0.2)

• Case 3: A misses ($1-0.4 = 0.6$), B hits (0.3), C hits (0.2) $0.6 \times 0.3 \times 0.2 = 0.036$ …

Method: Exactly "$k$ of $n$" independent events with different probabilities

Use this whenever several independent events each occur once, each with its own probability, and you want exactly a certain number of them to happen.

Steps

Step 1: Confirm independence and list the individual probabilities.

Check that no event's outcome changes another's probability. Write each success probability $p_i$ and its complement (failure) $q_i = 1 - p_i$. The multiplication theorem for independent events, $P(E_1 \cap E_2 \cap \dots) = P(E_1)P(E_2)\cdots$, is what lets you multiply.

Step 2: List every case that gives exactly the required number of successes.

For "exactly $k$ of $n$", choose which $k$ events succeed and which $n-k$ fail — there are $\binom{n}{k}$ such cases. Because the probabilities differ, you cannot collapse them into one term; keep them separate. …

Common Mistakes

Mistake 1: Multiplying all three hit probabilities, $0.4\times0.3\times0.2 = 0.024$.

Why it's wrong: that product is the probability of three hits (everyone hits), not two. Correct approach: exactly two hits requires exactly one miss, so include a miss factor $(1-p)$ for the person who misses.

Mistake 2: Forgetting the miss probability is $1-p$, not $p$.

Why it's wrong: using the hitter's probability for the person who misses double-counts hits. Correct approach: the misser contributes $1-p$ (e.g. C misses with $1-0.2 = 0.8$). …