An electronic product contains 32 integrated circuits. The probability that any integrated circuit is defective is 0.02, and the integrated circuits are independent. The product operates only if there are no defective integrated circuits. What is the probability that the product operates?
I know that it follows x~Binomial(32,.02)
I just can't figure out what they want me to find.
Its x~Bin(32,.98) since were looking for when it WILL operate and x = 32.
To find the probability that the product operates, we need to find the probability that none of the 32 integrated circuits are defective.
Since the probability of any integrated circuit being defective is 0.02, the probability of any integrated circuit not being defective (i.e., operating) is 1 - 0.02 = 0.98.
Since the integrated circuits are independent, we can use the binomial distribution to calculate the probability that none of the 32 integrated circuits are defective.
The formula for the binomial probability is P(X = k) = (n choose k) * p^k * (1 - p)^(n - k), where n is the number of trials, k is the number of successful outcomes, p is the probability of success, and (n choose k) is the combination of n and k.
In this case, n = 32 (the number of integrated circuits), k = 0 (the number of defective integrated circuits we want), and p = 0.98 (the probability of an integrated circuit operating).
Using the binomial probability formula, we can calculate the probability that none of the 32 integrated circuits are defective:
P(X = 0) = (32 choose 0) * (0.98^0) * (0.02^(32 - 0))
= 1 * 1 * (0.02^32)
= 1 * 1 * 0.0000000000000000000000123133546204313
≈ 0
Therefore, the probability that the product operates is approximately 0.