ProbabilityMedium
Question
From a lot containing 10 defective and 90 non-defective bulbs, 8 bulbs are selected one by one with replacement. Then the probability of getting at least 7 defective bulbs is :-
Options
A.$\frac{7}{10^{7}}$
B.$\frac{81}{10^{8}}$
C.$\frac{67}{10^{8}}$
D.$\frac{73}{10^{8}}$
Solution
10 defective & 90 non-defective
Req. probability $= (7$ def 1 fair $)$ or ( 8 defective)
Req. probability $= \frac{\left( 10^{7} \times 90 \right) \times 8 + 10^{8}}{100^{8}}$
$$= \frac{72 \times 10^{8} + 10^{8}}{100^{8}} = \frac{73}{10^{8}}$$
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