Two shooters independently of each other shoot once at the target with hitting probabilities of 0.6 and 0.7

Two shooters independently of each other shoot once at the target with hitting probabilities of 0.6 and 0.7, respectively. Determine the probability that only one of them will hit.

The probability that the first shooter will hit is p1 = 0.6.
The probability that the first shooter will not hit: q1 = 1 – p1 = 1 – 0.6 = 0.4.
The probability that the second shooter will hit is p2 = 0.7.
The probability that the first shooter will not hit: q2 = 1 – p2 = 1 – 0.7 = 0.3.
Let’s find the probability that only one shooter will hit.
P = p1 * q2 + q1 * p2 = 0.6 * 0.3 + 0.4 * 0.7 = 0.46.
Answer: The probability that only one shooter will hit is P = 0.46.