Question

A student needs eight chips of a certain type to build a circuit. It is known that 5% of these chips are defective. How many chips should he buy for there to be a greater than 90% probability of having enough chips for the circuit? (Use Binomial Probability Law and consider n = 8, 9, ...)

Answer 1

Binomial probability law:

$$P(k,n)=\frac{n!}{k!(n-k)!}p^k(1-p)^{n-k}$$

p=0.95.

To determine the probability that the working chips are greater than seven, we add the probabilities for each k starting from 8 and ending with n.

n=8: P=P(8,8)≈0.663;

n=9: P=P(8,9)+P(9,9)≈0.929.

Answer: n=9.