Question

A lot of 100 bulbs contains 10 defective bulbs. Five bulbs are selected at random from the lot and are sent to retail store. Then the probability that the store will receive at most one defective bulb is

• A. $\frac{7}{5} \left(\frac{9}{10}\right)^4$
• B. $\frac{7}{5} \left(\frac{9}{10}\right)^5$
• C. $\frac{6}{5} \left(\frac{9}{10}ight)^4$
• D. $\frac{6}{5} \left(\frac{9}{10} ight)^5$

Hint:

When dealing with 'at most one' scenarios in binomial probability, calculate the probabilities for $X=0$ and $X=1$ and sum them. Remember to correctly identify $n, p, q$ and use the combination formula ${^nC_r}$.

Answer 1

The Correct Option is A

Step 1: Define the random variable and probabilities. Let $X$ denote the number of defective bulbs in the selection of 5. This is a binomial distribution scenario since we are sampling with replacement (implicitly, or a large lot).


However, it's a hypergeometric distribution if sampling without replacement. Given the solution uses binomial, we proceed with that assumption, or a large enough lot that binomial approximation is valid for defective items.
Step 2: Calculate the probability of a defective bulb ($p$) and a non-defective bulb ($q$). Given that a lot of 100 bulbs contains 10 defective bulbs. Probability that a bulb is defective, $p = \frac{10}{100} = \frac{1}{10}$. Probability that a bulb is not defective, $q = 1 - p = 1 - \frac{1}{10} = \frac{9}{10}$.



Step 3: Set up the probability distribution for $X$. For a binomial distribution, the probability of getting $r$ defective bulbs out of 5 selected is given by: \[ P(X = r) = {^5C_r} p^r q^{5-r} \] \[ P(X = r) = {^5C_r} \left(\frac{1}{10}\right)^r \left(\frac{9}{10}\right)^{5-r}, \quad r = 0, 1, ......., 5 \]
Step 4: Calculate the probability of receiving at most one defective bulb. "At most one defective bulb" means $P(X \le 1)$, which is $P(X=0) + P(X=1)$. \[ P(X \le 1) = P(X = 0) + P(X = 1) \]
Step 5: Calculate $P(X=0)$. \[ P(X = 0) = {^5C_0} \left(\frac{1}{10}\right)^0 \left(\frac{9}{10}\right)^{5-0} \] \[ P(X = 0) = 1 \cdot 1 \cdot \left(\frac{9}{10}\right)^5 = \left(\frac{9}{10} ight)^5 \]
Step 6: Calculate $P(X=1)$. \[ P(X = 1) = {^5C_1} \left(\frac{1}{10}\right)^1 \left(\frac{9}{10}\right)^{5-1} \] \[ P(X = 1) = 5 \cdot \left(\frac{1}{10}\right) \cdot \left(\frac{9}{10}\right)^4 \]
Step 7: Sum the probabilities. \[ P(X \le 1) = \left(\frac{9}{10}\right)^5 + 5 \left(\frac{1}{10}\right) \left(\frac{9}{10} ight)^4 \] \[ P(X \le 1) = \left(\frac{9}{10} ight)^5 + \frac{5}{10} \left(\frac{9}{10} ight)^4 \] \[ P(X \le 1) = \left(\frac{9}{10} ight)^4 \left[ \left(\frac{9}{10}\right)^1 + \frac{5}{10} \right] \] \[ P(X \le 1) = \left(\frac{9}{10} ight)^4 \left[ \frac{9}{10} + \frac{5}{10} \right] \] \[ P(X \le 1) = \left(\frac{9}{10} ight)^4 \left[ \frac{14}{10} \right] \] \[ P(X \le 1) = \left(\frac{9}{10} ight)^4 \left[ \frac{7}{5} \right] \] \[ P(X \le 1) = \frac{7}{5} \left(\frac{9}{10} ight)^4 \]