Second question

**BaBrown1s** · May 7th 2009, 07:44 PM

The computer manufacturer plans to test 4 computers out of the shipment of 22 for testing. Suppose 2 of the 22 are faulty.

a. How many ways can he choose the four to test?

b. What are the chances that both of flawed computers are in the test group.

**pickslides** · May 7th 2009, 08:07 PM

Originally Posted by BaBrown1s

The computer manufacturer plans to test 4 computers out of the shipment of 22 for testing. Suppose 2 of the 22 are faulty.

a. How many ways can he choose the four to test?

b. What are the chances that both of flawed computers are in the test group.

a) $\binom{22}{4}= 7315$

b) use binomial theorem

$P(X=k) = \binom{n}{k}p^k(1-p)^{n-k}$

where n = 4, p= 2/22, k=2

$P(X=2) = \binom{4}{2}(\tfrac{2}{22})^2(1-\tfrac{2}{22})^{4-2}$

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