Statistics Q2

Method 1: Using the hints

P(X1 + X2 = 1)

=P(X1 = 0,X2 = 1) + P(X1 = 1,X2 = 0)

= (980/1000)(20/999) + (20/1000)(980/999)

= 39200/999000

= 196/4995

P(X1 + X2 = 2)

= (20/1000)(19/999)

= 380/999000

= 19/49950

The probability that at least one light bulb is defective

= 196/4995 + 19/49950

= 1979/49950

Method 2: Using hypergeometric distribution

The probability that at least one light bulb is defective

= 1 - P(X1 + X2 = 0)

= 1 - (980C2)/(1000C2)

= 1 - 479710/499500

= 19790/499500

= 1979/49950