Answer to Question #317308 in Statistics and Probability for angel

Question

Suppose three cellphones are tested at random. We want to find out the number of defective cellphones that occur. Let X be the random variable that represents the number of defective cellphones. Let D represent the defective cell phone and N for non-defective.

Answer 1

Let probability of N is equal to probability of D and equal 0.5

P(NNN)=(0.5)3=0.125

P(DDD)=0.125

P(NDD)=0.125

P(DND)=0.125

P(DDN)=0.125

P(DNN)=0.125

P(NDN)=0.125

P(NND)=0.125

$μ=3x0.125+0x0.125+3x1x0.125+3x2x0.125=1.5$