Suppose that 17 lightbulbs in a box of 100 lightbulbs are br

Suppose that 17 lightbulbs in a box of 100 lightbulbs are broken and that 3 are selected at random without replacement. Construct a probability tree for this problem. What is the probability that there will be no broken lightbulbs in the sample? What is the probability that there will be no more than 1 broken lightbulb in the sample? (This problem is continued in Problem 1.7.8.)

Solution

Binomial Distribution

PMF of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k
Where   
k = number of successes in trials
n = is the number of independent trials
p = probability of success on each trial

\'0\' bronken bulb in the sample
P( X = 0 ) = ( 3 0 ) * ( 0.17^0) * ( 1 - 0.17 )^3
= 0.5718

\'1\' bronken bulb in the sample
P( X = 1 ) = ( 3 1 ) * ( 0.17^1) * ( 1 - 0.17 )^2
= 0.3513

\'2\' bronken bulb in the sample
P( X = 2 ) = ( 3 2 ) * ( 0.17^2) * ( 1 - 0.17 )^1
= 0.072

\'3\' bronken bulb in the sample
P( X = 3 ) = ( 3 3 ) * ( 0.17^3) * ( 1 - 0.17 )^0
= 0.0049

Probability No broken bulb in the sample = P( X = 0 ) = 0.5718
Probability more than 1 broken in sample =
P( X < 1) = P(X=0)   
= ( 3 0 ) * 0.17^0 * ( 1- 0.17 ) ^3
= 0.5718
P( X > = 1 ) = 1 - P( X < 1) = 0.4282