A box contains 10 light bulbs and 2 of the 10 bulbs are defective. 3 of the 10 bulbs are taken randomly from the box. Find the probability that there is exactly 1 defective bulb out of the 3 bulbs taken from the box
Answers
7/15 is the probability that there is exactly 1 defective bulb out of the 3 bulbs taken from the box if 2 of the 10 bulbs are defective
Step-by-step explanation:
Total bulbs = 10
Defective = 2
non - Defective = 8
3 Bulbs selected and we need to find the probability that there is exactly 1 defective bulb
There can be 3 cases
1st bulb is defective
2nd bulb is defective
3rd bulb is defective
Probability 1st bulb is defective = (2/10)(8/9)(7/8) = 7/45
Probability 2nd bulb is defective = (8/10)(2/9)(7/8) = 7/45
Probability 3rd bulb is defective = (8/10)(7/9)(2/8) = 7/45
Total probability = 7/45 + 7/45 + 7/45 = 7/15
Another way
P(1) = ³C₁(2/10)¹(8/10)² = 12/25 ( but it works where Volume is high & Probability of defective & non-defective remains constant)
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