Question

По одному и тому же маршруту в один и тот же день совершают полет 3 самолета. Каждый самолет с вер-тью 0.7 может произвести посадку по расписанию. Какова вероятность, что 3 самолета произведут посадку по расписанию? Не более двух самолетов?

Answer 1

Для трех самолетов получаем по теореме умножения
Р(3)=0,7*0,7*0,7=0,343
не более 2, значит 0, 1 или 2, находим через вероятность противоположного события Р=1-Р(3)=1-0,343=0,657

Answer 2

I will answer in detail. The probability of a certain event is the ratio of the number of favorable outcomes to the total number of possible outcomes. In this case, the event is that 3 planes land on schedule, and the probability of each plane landing on schedule is 0.7. The total number of possible outcomes is 2^3 = 8, since each plane can either land on schedule or not. The number of favorable outcomes is 1, since only one combination of outcomes satisfies the event: all three planes land on schedule. Therefore, the probability of the event is 1/8 = 0.125.

The probability of another event is the sum of the probabilities of all the outcomes that satisfy the event. In this case, the event is that no more than two planes land on schedule. The outcomes that satisfy the event are: two planes land on schedule and one does not, or one plane lands on schedule and two do not, or none of the planes land on schedule. The probability of each of these outcomes is 0.7^2 * 0.3, 0.7 * 0.3^2, and 0.3^3, respectively. Therefore, the probability of the event is 0.7^2 * 0.3 * 3 + 0.7 * 0.3^2 * 3 + 0.3^3 = 0.783.