Question ID: #867
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
- (1) $\frac{7}{10^{7}}$
- (2) $\frac{81}{10^{8}}$
- (3) $\frac{67}{10^{8}}$
- (4) $\frac{73}{10^{8}}$
Solution:
Let $p$ be the probability of selecting a defective bulb.
$$p = \frac{10}{100} = \frac{1}{10} = 0.1$$
Let $q$ be the probability of selecting a non-defective bulb.
$$q = 1 – p = 0.9$$
Let $X$ be the random variable representing the number of defective bulbs in 8 trials.
We need to find $P(X \ge 7)$, which means getting 7 or 8 defective bulbs.
$$P(X \ge 7) = P(X=7) + P(X=8)$$
Using the Binomial Probability formula $P(X=r) = {^nC_r} p^r q^{n-r}$:
$$P(X=7) = {^8C_7} (0.1)^7 (0.9)^1$$
$$P(X=8) = {^8C_8} (0.1)^8 (0.9)^0$$
Substitute the values:
$$P(X \ge 7) = 8 \cdot \frac{1}{10^7} \cdot \frac{9}{10} + 1 \cdot \frac{1}{10^8} \cdot 1$$
$$P(X \ge 7) = \frac{72}{10^8} + \frac{1}{10^8}$$
$$P(X \ge 7) = \frac{73}{10^8}$$
Ans. (4)
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