Three light bulbs are chosen at random from 15 bulbs of which 5 are defective. Find the probability p that atleast one is defeactive
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Total bulbs = 15
Defective = 5
P(E) defective one = 5/15 = 1/3
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This is a method in which you will be able to solve probability problems without using permutations and combinations
A : bulb is not defective
B : bulb is defective
P(A) = 10/15 = 2/3
P(B) = 5/15 = 1/3
=3P(ABB) + 3P(AAB) + 3P(BBB) ....( since each case occurs thrice)
=3(1/3 . 1/3 . 2/3) + 3(1/3 .2/3. 2/3) + 3(1/3. 1/3. 1/3)
=2/9 + 4/9 + 1/9
=7/9
Step-by-step explanation:
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