Both Torpedoes Hit Given Target Sunk
A submarine fires two torpedoes at a target. Each torpedo independently hits with probability $p = 3/7$. Assume the target sinks if and only if at least one torpedo hits.
Given that the target has sunk, what is the probability that both torpedoes hit?
Hints
- The target sinking is equivalent to at least one torpedo hitting. You want $P(\text{both hit} \mid \text{at least one hit})$. This is a conditional probability -- write it using the definition.
- Use $P(A \cap B \mid A \cup B) = P(A \cap B) / P(A \cup B)$, since $A \cap B \subseteq A \cup B$. Compute each piece using independence.
- $P(A \cap B) = (3/7)^2 = 9/49$. $P(A \cup B) = 1 - (4/7)^2 = 33/49$. Divide to get $9/33 = 3/11$.
Worked Solution
How to Think About It: You are doing a Bayesian update: the target sinking is evidence that at least one torpedo hit. Given that at least one hit, what are the chances both hit? Think of it as conditioning on the union (at least one hit) and asking for the probability of the intersection (both hit). This is a direct application of Bayes' theorem or conditional probability.
Quick Estimate: Each torpedo hits with probability $3/7 \approx 43\%$. The probability both hit is $(3/7)^2 = 9/49 \approx 18\%$. The probability at least one hits is
Approach: Apply conditional probability directly. The event "target sunk" is equivalent to "at least one torpedo hit."
Formal Solution:
Let $A$ = torpedo 1 hits, $B$ = torpedo 2 hits. Both are independent with $P(A) = P(B) = 3/7$.
The target sinks iff at least one hits: $\{\text{sunk}\} = A \cup B$.
We want $P(A \cap B \mid A \cup B)$.
By conditional probability: $P(A \cap B \mid A \cup B) = \frac{P(A \cap B \cap (A \cup B))}{P(A \cup B)} = \frac{P(A \cap B)}{P(A \cup B)}$
since $A \cap B \subseteq A \cup B$ (both hitting implies at least one hit).
Numerator: $P(A \cap B) = P(A) \cdot P(B) = \frac{3}{7} \cdot \frac{3}{7} = \frac{9}{49}$
Denominator: $P(A \cup B) = 1 - P(A^c \cap B^c) = 1 - \frac{4}{7} \cdot \frac{4}{7} = 1 - \frac{16}{49} = \frac{33}{49}$
Result: $P(A \cap B \mid A \cup B) = \frac{9/49}{33/49} = \frac{9}{33} = \boxed{\frac{3}{11}} \approx 27.3\%$
This matches the quick estimate.
Intuition
This is a clean Bayesian update: observing that the target sank increases the probability that both torpedoes hit (relative to the unconditional probability). Before observing the outcome, $P(\text{both hit}) = 9/49 \approx 18\%$. After observing the target sank (so at least one hit), $P(\text{both hit} \mid \text{sunk}) = 3/11 \approx 27\%$. The conditioning event is informative -- learning that the target sank rules out the "both miss" scenario, which reweights the remaining probability mass upward toward the "both hit" state.
The general formula here is $P(A \cap B \mid A \cup B) = P(A \cap B) / P(A \cup B)$, which follows from the fact that $\{A \cap B\}$ is a subset of $\{A \cup B\}$. This structure -- conditional probability as re-normalization over a restricted sample space -- is the engine behind many market microstructure models, where observing a trade or price move updates the posterior probability of the underlying information state.