SOLUTION: Okay This was a two part question I got the second part correct but cannot seem to get the first part of the question correct.
Defective Refrigerator Through a mix-up on the pro
Algebra.Com
Question 192571This question is from textbook Finite Mathematics
: Okay This was a two part question I got the second part correct but cannot seem to get the first part of the question correct.
Defective Refrigerator Through a mix-up on the production line, 6 defective refrigerators were shipped out with 39 good ones.
1) If 5 are selected at random, what is the probability that all 5 are defective? Round your answer to 6 decimal places.
2) What is the probability that at least 2 of them are defective? Round your answer to 3 decimal places. ANS = 0.125
This question is from textbook Finite Mathematics
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
The first part is a conditional probability problem. Since there are 6 defective refrigerators in a batch of 39, and you select 1 of them, then the probability that you will have selected a defective one is 
.
Now, given that you were successful and selected a defective unit on the first try, and making the entirely reasonable assumption that, having found a bad one, you wouldn't put it back, when you go to select the second random unit, there are only 5 defective ones remaining out of 38 total. (1 less defective one and 1 less total to choose from). Hence, the probability on the second selection is 
.
Continuing in that vein, each time presuming that you were successful on the previous trial, the remaining probabilities are:
And finally, the total probability of 5 defective out of 5 selected from 39 is the product of those 5 probabilities, or:
You can do the arithmetic.
John
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