Question

The probability of an integrated circuit passing a quality test is 0.86. In a batch of 10 integrated circuits find the probabilities that
2 will fail
8 will pass
at least 2 will fail
fewer than 2 will fail.

Answer 1

Let $X=$ the number of integrated circuits which will pass a quality test: $X\sim Bin(n, p).$

Given $n=10, p=0.86, q=1-p=1-0.86=0.14.$

a)

$$P( \text{2 will fail})=P(X=8)$$

$$=\dbinom{10}{8}(0.86)^8(0.14)^{10-2}=0.26391$$

b)

$$P( \text{8 will pass})=P(X=8)$$

$$=\dbinom{10}{8}(0.86)^8(0.14)^{10-2}=0.26391$$

c)

$$P( \text{at least 2 will fall})=P(X\leq8)$$

$$=1-(P(X=9)+P(X=10))$$

$$=1-\dbinom{10}{9}(0.86)^9(0.14)^{10-9}$$

$$-\dbinom{10}{10}(0.86)^{10}(0.14)^{10-10}$$

$$=0.41844$$

d)

$$P( \text{fewer than 2 will fail})=P(X>8)$$

$$=P(X=9)+P(X=10)$$

$$=\dbinom{10}{9}(0.86)^9(0.14)^{10-9}+\dbinom{10}{10}(0.86)^{10}(0.14)^{10-10}$$

$$=0.58156$$