Mathematical probability shooting problem

Whether three people hit the target is an independent event. The event probability of hitting a is PA=0.7, that of hitting b is PB=0.6, that of hitting c is PC=0.8, and that of three people hitting at the same time is P (a, b, c) = PA * Pb * PC = 0.7 * 0.6 * 0.8 = 0.38.

The event probability that three people only hit the target twice = pa * Pb * (1-PC)+pa * PC (1-Pb)+PC * Pb * (1-pa) = 0.7 * 0.6 * 0.2+0.7 * 0.8. Probability is the sum of the probabilities of these three events. One of the three events A, B and C is not equal, and three independent events A, B and C miss at the same time. The probability of this event is the sum of three events, where (1-PA) refers to the probability of miss, and the rest are the same).

Event probability of three people hitting the target only once = pa * (1-Pb) * (1-PC)+Pb * (1-pa) * (1-PC)+PC *.

The incident probability of missing three people = (1-pa) * (1-Pb) * (1-PC) = 0.3 * 0.4 * 0.2 = 0.024.

P All events = 0.024+0. 1 88+0.452+0.336 =1(there are only four possible shooting results, and the probability sum is1).

After the answer, the main concept is that the probability of several independent events happening at the same time is the product of the probability of each event (the so-called independence means that they do not affect each other, event A does not happen, nor does it affect the occurrence of event B, and vice versa). When an event is the sum of several mutually exclusive events (that is, any event such as A or B can happen, but B will not happen when A happens, and vice versa, A and B are mutually exclusive events), its probability.

I hope it helps you.