Answer:
P(X=0)=0.732
P(X=1)=0.1962
P(X=2)=0.0526
P(X=3)=0.0141
P(X=4)=0.0038
P(X=5)=0.001 0
P(X≥6)=0.0004
Explanation:
We want to find the probability distribution of X, where X is the number of items that must be tested before finding the first defective 10-year old widget.
This means that if X=2, the first two items are not defective and the third item is defective.
The probability of finding a defective item is p=0.732 for each trial.
Then, the probability for X=k can be written as:

Then, we can write:
