An enemy aircraft can be shot down by each of three anti-aircraft guns with probabilities 0.2,0.3,0.4

Question

An enemy aircraft can be shot down by each of three anti-aircraft guns with probabilities $0.2,0.3,0.4$

1. what is the probability that the aircraft will be shot down with the simultaneous firing of all guns?
2. What is the probability that it will be shown down with repeated firing if the first one unsuccessful?
3. the anti-aircraft guns were successful in the first simultaneous firing .what is the likelihood that the third gun was successful was shot down by the third gun if it is known that only one of the guns hit the target?
4. what is the probability that it was shot down by the third gun if it is known that only of the two guns hit the target?
   

my try

1. $0.9$
2. $1-(0.8*0.7*0.6)$
3. idk
4. $(1/2)*0.4$

Answer 1

The aircraft will be shot down if at least one of the anti-aircraft guns hits the enemy aircraft. The probability is equal to 1 minus the probability that no anti-aircraft gun hits the enemy aircraft: $1-(1-0.2)\cdot (1-0.3)\cdot (1-0.4)=1-0.8\cdot 0.7\cdot 0.6=66.4\%$. So your answer for $2.$ is the answer for $1.$

Second question:

I assume that two shots are made. The aircraft is not hit by the first shot of the three guns but by the second shot by at least one of the 3 guns. The probability is

$(1-0.2)\cdot (1-0.3)\cdot (1-0.4)\cdot (1-(1-0.2)\cdot (1-0.3)\cdot (1-0.4))$

$1-(1-0.2)\cdot (1-0.3)\cdot (1-0.4)$: This is the probability that at least one of the 3 guns hit the aircraft.

Third question:

Suppose we denote the event that the aircraft is hit by gun i as $h_i$. The converse event is denoted by $\overline h_i$

The 3 events for only one of the guns hit the target are:

\begin{aligned} \\ h_1 \overline h_2 \overline h_3 \\ \overline h_1 h_2 \overline h_3 \\ \overline h_1 \overline h_2 h_3 \end{aligned}

Now we identifiy the event that the third gun was successful. This is $\overline h_1 \overline h_2 h_3$

Thus the asked probability is $$\frac{P(\overline h_1 \overline h_2 h_3)}{P(h_1 \overline h_2 \overline h_3)+P(\overline h_1 h_2 \overline h_3)+P(\overline h_1 \overline h_2 h_3)}=\frac{0.8*0.7*0.4}{0.2*0.7*0.6+0.8*0.3*0.6+0.8*0.7*0.4}=$$

$=0.495575\approx 49.56\%$

Fourth question:

Same procedure as in $3$.

\begin{aligned} \\ h_1 h_2 \overline h_3 \\ h_1 \overline h_2 h_3 \\ \overline h_1 h_2 h_3 \end{aligned}

$$ \frac{P(h_1 \overline h_2 h_3)+P(\overline h_1 h_2 h_3)}{P(h_1 h_2 \overline h_3)+P(h_1 \overline h_2 h_3)+P(\overline h_1 h_2 h_3)}=\frac{0.2*0.7*0.4+0.8*0.3*0.4}{0.2*0.3*0.6+0.2*0.7*0.4+0.8*0.3*0.4}\approx 80.85\%$$

Answer 2

Assuming prob. of hits be each gun independent of others.

1) $.664$ = (1 - prob all miss).

2) $1$ - if they keep on firing eventually there will be a hit.

3) $0.495575221238938018$ = (prob hit by 3 and other two miss)/(prob exactly one hit).

4) $0.808510638297872271$ = (prob hit by 3 and one other)/(prob exactly two hit).