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 [2026]

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 [2026]

Smart Achievers · Accepted Answer

(1)

$10 defective & 90 non-defective$

$Req. probability = (7 defective & 1 fair) or (8 defective)$

$Req. probability = \frac{(10^{7} \times 90) \times 8 + 10^{8}}{100^{8}}$

$= \frac{72 \times 10^{8} + 10^{8}}{100^{8}} = \frac{73}{10^{8}}$

Solution

Q.
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 [2026]

1 $\frac{73}{10^{8}}$

2 $\frac{67}{10^{8}}$

3 $\frac{81}{10^{8}}$

4 $\frac{7}{10^{7}}$

Ans.
(1)

$10 defective & 90 non-defective$

$Req. probability = (7 defective & 1 fair) or (8 defective)$

$Req. probability = \frac{(10^{7} \times 90) \times 8 + 10^{8}}{100^{8}}$

$= \frac{72 \times 10^{8} + 10^{8}}{100^{8}} = \frac{73}{10^{8}}$

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Solution

Q. 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 [2026]

1 73108

2 67108

3 81108

4 7107

Ans. (1) 10 defective & 90 non-defective Req. probability=(7 defective & 1 fair) or (8 defective) Req. probability=(107×90)×8+1081008 =72×108+1081008=73108