if 6 electronic components are tested and each has a 03 prob

if 6 electronic components are tested and each has a 0.3 probability of being defective, what is the probability that the first 3 components will be tested to be ok before a defective component is discovered?

Solution

given n=6

probabiility of defective=0.3

prob of non defective=0.7

There are two ways to interpret your question:
1. At least 3 components are ok before a defective one is found (sometime)
2. Exactly 3 components are ok before the 4th is found to be defective

For 1 the answer is (0.7)³ = 0.343
For 2 the answer is (0.7)³ 0.3 = 0.1029.
In the latter we need to multiply by the probability of the 4th component being defective!