Answer to Question #154558 in Statistics and Probability for 75ed86

Question #154558

A box contains ten light bulbs of which four are defective. A bulb is selected from the box and tested. If it is defective, another bulb is selected and tested, until a non-defective bulb is chosen. Let the random variable X be the number of bulbs chosen. 

a) Write sample space for this random experiment?

b) Find probability mass Function of X? 

c) Find CDF of X?


1
Expert's answer
2021-01-11T18:08:50-0500

a) Write sample space for this random experiment?

RX = {1,2,3,4,5,6,7}

b) Find probability mass Function of X? 

P(X=1)=410=0.4P(X=2)=61049=4/15=0.2667P(X=3)=6105948=1/6=0.1667P(X=4)=610594847=2/21=0.0952P(X=5)=61059483746=1/21=0.0476P(X=6)=6105948372645=2/105=0.0190P(X=7)=610594837261544=1/210=0.048P(X=1) = \frac4{10}=0.4\\ P(X=2) = \frac6{10}\frac49 = 4/15=0.2667\\ P(X=3) = \frac6{10}\frac59\frac48=1/6=0.1667\\ P(X=4) = \frac6{10}\frac59\frac48\frac47=2/21=0.0952\\ P(X=5) = \frac6{10}\frac59\frac48\frac37\frac46=1/21=0.0476\\ P(X=6) = \frac6{10}\frac59\frac48\frac37\frac26\frac45=2/105=0.0190\\ P(X=7) = \frac6{10}\frac59\frac48\frac37\frac26\frac15\frac44=1/210=0.048

c) Find CDF of X?

P(X1)=0.4P(X2)=2/5+4/15=2/3=0.6667P(X3)=2/3+1/6=5/6=0.8333P(X4)=5/6+2/21=13/14=0.9286P(X5)=13/14+1/21=41/42=0.9762P(X6)=20/21+2/105=209/210=0.9952P(X7)=1.0P(X\le1) =0.4\\ P(X\le2) =2/5+4/15=2/3=0.6667\\ P(X\le3) = 2/3+1/6=5/6=0.8333\\ P(X\le4) = 5/6+2/21=13/14=0.9286\\ P(X\le5) = 13/14+1/21=41/42=0.9762\\ P(X\le6) = 20/21+2/105=209/210=0.9952\\ P(X\le7) = 1.0


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