On average 3 % of a certain type of TV tube burn out before their guarantee has expired. If a merchant sells 120 such tubes, find the probability that he will be forced to replace at least 3 of them (Use the Poisson approximation to the Binomial distribution).
Never mind, got it!!
To find the probability of replacing at least 3 TV tubes, we first need to calculate the probability of replacing exactly 0,1, and 2 tubes, and then subtract that from 1 to get the probability of replacing at least 3 tubes.
To do this, we can use the Poisson approximation to the Binomial distribution. The Poisson distribution is often used as an approximation for the Binomial distribution when the number of trials is large and the probability of success is small.
For a given TV tube, the probability of it burning out before the guarantee expires is 3%. This means the probability of not burning out before the guarantee expires is 1 - 3% = 97% = 0.97.
Let's define the following variables for calculations:
n = number of trials (120 TV tubes)
p = probability of success (0.03, since 3% burn out before the guarantee expires)
The mean of the Poisson distribution is given by λ = n * p = 120 * 0.03 = 3.6.
Now we can calculate the probability of replacing exactly k tubes using the Poisson distribution formula:
P(X = k) = (e^(-λ) * λ^k) / k!
For k = 0: P(X = 0) = (e^(-3.6) * 3.6^0) / 0! = e^(-3.6)
For k = 1: P(X = 1) = (e^(-3.6) * 3.6^1) / 1! = e^(-3.6) * 3.6
For k = 2: P(X = 2) = (e^(-3.6) * 3.6^2) / 2! = e^(-3.6) * (3.6^2 / 2)
To find the probability of replacing at least 3 tubes, we need to subtract the probabilities of replacing 0, 1, and 2 tubes from 1:
P(X ≥ 3) = 1 - P(X = 0) - P(X = 1) - P(X = 2)
= 1 - e^(-3.6) - e^(-3.6) * 3.6 - e^(-3.6) * (3.6^2 / 2)
Calculating this value will give us the probability that the merchant will be forced to replace at least 3 TV tubes.