Please, can someone help me with this question. I need to understand the steps or reasoning to arrive at the answer and not just the answer. A company that produces fine crystal knows from experience that 10% of its goblets has flaws and must be classified as 'seconds'. If goblets are examined one by one, what is the probability that at most five must be selected to find four that are not 'seconds'. Here is how I went about it. I analyzed this as a binomial distribution and then said: X = number not seconds. p= 0.9 x = 4 Then what varies is n, i.e what n gives the probability for P(X=4). I then told myself that to give 4 non-seconds, then we need to examine either 4 goblets (all are not seconds) or five goblets (4 are not seconds and 1 is a second). So, this is the addition of two binomial probability mass functions, b(4; 4, 0.9) + b(4; 5, 0.9)
b(4; 4, 0.9) = $ {4 \choose 4} (0.9)^4 (0.1)^0$ = 0.6561
b(4; 5, 0.9) = ${5 \choose 4}(0.9)^4(0.1) $ = 0.32805
the total gives = 0.9842.
But the textbook says that the answer is 0.918. Please, can someone explain to me where I went wrong and how I should have tackled the problem. Or what principle in the problem I am missing? thanks
You are unlucky...if they asked you "at most 6", for example, with your method you found a probability of about 1.08....greater than 1. So you would have realized that your method is wrong, also without reading the correct solution.
The correct solution is to use a Negative Binomial, the distribution that counts how many trials are necessary to get a fixed number of successes. In your cases
$$\binom{4-1}{4-1}0.9^4\cdot0.1^{4-4}+\binom{5-1}{4-1}0.9^4\cdot0.1^{5-4}=0.656+0.262=0.91854\approx0.919$$