QUESTION 18 MATH NEED ASSISTANCE ASAP PLEASE

Question

a box contains 14 trasistors, 3 of which are defective, if 3 are selected at random, find the probability of the statments below.

a. all are defective

b. none are defective

Question 1. The probability is .....

Question 2 The Probability is......

Answer 1

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 (rounded)

Answer to part b): 0.484 (rounded)