The probability that three computers need maintenance is:
p(3)=c(3,400)x0.0 1^3x( 1-0.0 1)^(400-3)=0. 1959
Similarly, p (2) = c (2400) x 0.012x (1-0.0/) 398 = 0.146.
p( 1)=c( 1,400)x0.0 1x0.99^399=0.0725
P(0)=0.99^400=0.0 1795
Therefore, the probability that three sets need maintenance is 0. 1959, and the probability that less than three sets (excluding three sets) need maintenance is 0.2365.
The probability that there are n cells to be repaired is: p (n) = c (n, 400) x 0.0 1 nx0.99 (400-n).
If there is an extreme value of n in this formula, calculate P(n)/P(n- 1), and it can be concluded that when n=(400+ 1)X0.0 1, P(n)=P(n- 1).
At this time, P(n) reaches the extreme point, and then with the increase of K value, P value will gradually decrease.
Therefore, n = (400+1) x 0.01= 4.1,which means that the maximum number of computers that can be repaired is 4, and the probability is p(4)= c(4400)x 0.0 1 4x 0.99.
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