Find the probability of these Computer Chips

Question

Computer Chips are manufactured with 20% defective. Five chips are drawn at random with replacement off an assembly line.

 

 

a. Find the probability that exactly 3 chips are defective.

b. Find the probability that at least one chip is defective.

c. Find the probability that at most one chip is defective.

Answer 1

Naively, this is the Binomial Distribution, where finding a defective chip is the event of interest, so is labeled as a "SUCCESS". The sampling must, as stated in the problem, be done with replacement.

 

The formula you need for finding "k successes in N trials" or in this case 3 defective chips from a sample of

5 selected randomly with replacement is:

  (N choose k) p^k (1-p)^(N-k) where N is the sample size, in this case 5,

                                                        p is the probability of "SUCCESS", which in this case is p=0.2 as given,

                                                        and (N choose k) = N!/{k!(N-k)!}

 

Expanding the factorials for this combinatoric results in:

N choose k = 5!/{3! (5-3!} = 5!/(3!2!) = (5*4*3*2*1)/{(3*2*1)(2*1)}

                                                       = (5*4)/(2*1)   <-- 3*2*1 = 6 cancels out

                                                        = 20/2

                                                        = 10

So the probability for part (a) is 10*(0.2)^3*(0.8)^2 = 10 * 0.2 * 0.2 * 0.2 * 0.8 * 0.8 = .0512 or just over 5%

 

Using this formula with these parameters and an excel spreadsheet, the following table shows the probability of getting k=0,1,2,3,4,5 success (defective computer chips). As a side note, the excel spreadsheet function COMBIN calculates the combinatoric part of the formula.

 

K P(N=k)
0 0.32768
1 0.4096
2 0.2048
3 0.0512
4 0.0064
5 0.00032 

 

So for part B, the probability of at least one defective chip is the same as the probability of ONE, TWO, THREE,

FOUR, or even FIVE defective chips in the sample. Adding the last probabilities in the table results in:

  0.4096 + 0.2048 + 0.0512 + 0.0064 + 0.00032 = 0.67232 or just over 67%

 

For part C, the probability of at most one chip is the same as ZERO or ONE defective chip in the same of five,

 which is 0.32768 + .4096 = 0.73728 or almost 74%