The probability of hitting the target by one shooter with one shot for the first shooter is 0.8

The probability of hitting the target by one shooter with one shot for the first shooter is 0.8, for the second shooter – 0.9, for the third shooter – 0.7. The shooters fired one shot at the target. Considering hitting the target for individual shooters as events independent, find the probability of event A – only one shooter hit.

Let us introduce the notation:

event E1 – the first shooter fired a successful shot;

event E2 – the second shooter fired a successful shot;

event E3 – the third shooter fired a lucky shot.

Let’s write down the probabilities of events E1, E2 and E3:

P (E1) = 0.8;

P (E2) = 0.9;

P (E3) = 0.7.

Let’s find the probabilities of opposite events.

P (Ē1) = 1 – P (E1) = 1 – 0.8 = 0.2;

P (Ē2) = 1 – P (E2) = 1 – 0.9 = 0.1;

P (Ē3) = 1 – P (E3) = 1 – 0.7 = 0.3.

Let’s say event A happened – only one of the three shooters hit the target. There are three options:

1. only the first shooter fired a successful shot;

2. only the second shooter fired a successful shot;

3. only the third shooter fired a good shot.

Find the probability that the first shooter fired a good shot and the other shooters missed.

P (E1) * P (Ē2) * P (Ē3) = 0.8 * 0.1 * 0.3 = 0.024.

Find the probability that the second shooter fired a good shot and the other shooters missed.

P (Ē1) * P (E2) * P (Ē3) = 0.2 * 0.9 * 0.3 = 0.054.

Find the probability that the third shooter fired a good shot and the other shooters missed.

P (Ē1) * P (Ē2) * P (E3) = 0.2 * 0.1 * 0.7 = 0.014.

Let’s find the probability that only one of the three shooters hit the target. To do this, add up the previously found probabilities.

P (A) = 0.024 + 0.054 + 0.014 = 0.092 = 9.2%.

So the probability that only one shooter hits the target is 9.2%.

Answer: 9.2 percent.