An air duel takes place between a fighter and a bomber. The…

An air duel takes place between a fighter and a bomber. The fighter fires first, one burst, and shoots down the bomber with probability $0.2$. If the bomber is not shot down, it returns fire and shoots down the fighter with probability $0.3$. If the fighter is not shot down, it attacks again from closer range and shoots down the bomber with probability $0.4$. Find the probability that in the duel $1^{\circ}$ the bomber is shot down; $2^{\circ}$ the fighter is shot down.

Explanation of Method

The duel is a sequence of at most three stages, and each stage happens only if the previous ones had a certain result. We follow the tree of possibilities: the probability of a path is the product of the conditional probabilities along it (multiplication rule), and the probability of an outcome is the sum over the paths leading to it (addition rule).

Step 1: The tree

Stage	Happens if	Result	Path probability
1: fighter fires	always	bomber down	$0.2$
2: bomber fires	bomber survived stage 1	fighter down	$0.8\cdot0.3=0.24$
3: fighter fires again	both survived	bomber down	$0.8\cdot0.7\cdot0.4=0.224$
end		both survive	$0.8\cdot0.7\cdot0.6=0.336$

Check: $0.2+0.24+0.224+0.336=1$ ✓.

Step 2: $1^{\circ}$ — the bomber is shot down

At stage 1 or at stage 3 (mutually exclusive):

\[ \boxed{P_1=0.2+0.8\cdot0.7\cdot0.4=0.2+0.224=0.424.} \]

Step 3: $2^{\circ}$ — the fighter is shot down

Only at stage 2:

\[ \boxed{P_2=0.8\cdot0.3=0.24.} \]

Step 4: Remark

Both aircraft cannot be shot down in this duel, and with probability $0.336$ neither is. Firing first gives the fighter a big advantage: even with a modest $0.2$ for its first burst, it wins more often than it loses.

Result

Part	Probability
$1^{\circ}$ bomber shot down	$0.2+0.8\cdot0.7\cdot0.4=0.424$
$2^{\circ}$ fighter shot down	$0.8\cdot0.3=0.24$