Question

A large shipment of light bulbs has just arrived at a store. It has been revealed that 17 % of the light bulbs are defective (the other light bulbs are good). Suppose that you choose 6 light bulbs at random. What is the probability that 2 or less of the bulbs are defective.?

Answer 1

The required probability = 0.9345

Explanation:

Given that:

p = 0.17

n = 6

The required probability that two or less light bulbs are defective = (P<2)

(P<2) = P(X=0) +P(X=1) +P(X=2)

`(P<2) =\bigg [ ({^6C_0}* 0.17^0 * (1-0.17)^(6-0))+ ({^6C_1}* 0.17^1 * (1-0.17)^(6-1)) +( {^6C_2}* 0.17^2 * (1-0.17)^(6-2)) \bigg ]`

`(P<2) =\bigg [ (6!)/(0!(6-0)!)* 0.17^0 * (1-0.17)^(6))+ ((6!)/(1!(6-1)!) * 0.17^1 * (1-0.17)^(5)) +( (6!)/(2!(6-2)! )* 0.17^2 * (1-0.17)^(4)) \bigg ]`

`\mathbf{(P<2) = 0.9345}`