Question

Among every 8 units of a product, one is likely to be defective. If a consumer has ordered 5 units of that product, then the probability that atmost one unit is defective among them is \

• A. \( \frac{15}{8} (\frac{7}{8})^6 \)
• B. \( \frac{57}{8^8} \)
• C. \( \frac{36}{8^5} \)
• D. \( \frac{3}{2} (\frac{7}{8})^4 \)

Hint:

In binomial probability, "at most k" means \( P(X=0) + P(X=1) + ... + P(X=k) \). "At least k" means \( P(X=k) + P(X=k+1) + ... + P(X=n) \). Recognizing these phrases is key to setting up the correct sum of probabilities.

Answer 1

The Correct Option is D

This is a binomial probability problem.
The probability of a single unit being defective is \( p = 1/8 \).
The probability of a single unit being non-defective is \( q = 1 - p = 1 - 1/8 = 7/8 \).
The number of trials (units ordered) is \( n = 5 \).
We need to find the probability that *at most one* unit is defective. This means either 0 units are defective OR 1 unit is defective.
P(at most 1 defective) = P(0 defective) + P(1 defective).
The binomial probability formula is \( P(X=k) = {^nC_k} p^k q^{n-k} \).
P(0 defective) = \( {^5C_0} (1/8)^0 (7/8)^{5-0} = 1 \cdot 1 \cdot (7/8)^5 = (7/8)^5 \).
P(1 defective) = \( {^5C_1} (1/8)^1 (7/8)^{5-1} = 5 \cdot (1/8) \cdot (7/8)^4 = \frac{5}{8} (\frac{7}{8})^4 \).
Now, add the two probabilities.
P(at most 1) = \( (\frac{7}{8})^5 + \frac{5}{8} (\frac{7}{8})^4 \).
To add these, let's factor out the common term \( (\frac{7}{8})^4 \).
P(at most 1) = \( (\frac{7}{8})^4 (\frac{7}{8} + \frac{5}{8}) \).
P(at most 1) = \( (\frac{7}{8})^4 (\frac{12}{8}) \).
Simplify the fraction \( \frac{12}{8} = \frac{3}{2} \).
So, the final probability is \( \frac{3}{2} (\frac{7}{8})^4 \).