SOLUTION: A box contains 15 transistors, 3 of which are defective. If 3 are selected at&#8203; random, find the probability that... a. All are defective. b. None are defective.

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Question 1039944: A box contains 15 transistors, 3 of which are defective. If 3 are selected at random, find the probability that...
a. All are defective.
b. None are defective.
Answer by jim_thompson5910(35256) (Show Source):
You can put this solution on YOUR website!

Part a)

There is only one way to select a batch of 3 transistors where they're all defective.

There are 15 C 3 = (15!)/(3!*(15-3)!) = 455 ways to select 3 resistors. This is where we don't care if they work or not.

The notation 15 C 3 refers to the combination formula

Dividing the two values gives 1/455 = 0.0021978021978. When we round to 3 decimal places, we get 0.002

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Part b)

There are 15-3 = 12 non-defective transistors. There are 12 C 3 = (12!)/(3!*(12-3)!) = 220 ways to pick 3 transistors where none of them are defective.

This is still out of the same total of 455 different combos (found earlier in part a)

220/455 = 0.48351648351649 which rounds to 0.484 (assuming your book wants you to round to 3 decimal places)

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Summary:
Answer to part a): 0.002
Answer to part b): 0.484

Answers are approximate. The values are rounded to 3 decimal places.

SOLUTION: A box contains 15 transistors, 3 of which are defective. If 3 are selected at&#8203; random, find the probability that... a. All are defective. b. None are defective.

SOLUTION: A box contains 15 transistors, 3 of which are defective. If 3 are selected at random, find the probability that... a. All are defective. b. None are defective.