Question

The probability that a certain kind of component will survive a given test is \(\frac{2}{3}\). The probability that at most \(2\) components out of \(4\) tested, will survive is

• A. \(\frac{31}{3^4}\)
• B. \(\frac{32}{3^4}\)
• C. \(\frac{33}{3^4}\)
• D. \(\frac{35}{3^4}\)

Hint:

“At most \(k\)” means include all cases from \(0\) to \(k\). That is the most common place where mistakes happen in binomial probability.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This problem follows a Binomial Distribution where each test is independent. Let \(X\) be the number of components surviving.
Step 2: Key Formula or Approach:
Binomial probability: \(P(X = r) = \binom{n}{r} p^r q^{n-r}\).
Here \(n = 4\), \(p = 2/3\), \(q = 1 - 2/3 = 1/3\).
At most \(2\) means \(P(X \le 2) = P(0) + P(1) + P(2)\).
Step 3: Detailed Explanation:
\[ P(0) = \binom{4}{0} \left(\frac{2}{3}\right)^0 \left(\frac{1}{3}\right)^4 = 1 \times 1 \times \frac{1}{81} = \frac{1}{81} \] \[ P(1) = \binom{4}{1} \left(\frac{2}{3}\right)^1 \left(\frac{1}{3}\right)^3 = 4 \times \frac{2}{3} \times \frac{1}{27} = \frac{8}{81} \] \[ P(2) = \binom{4}{2} \left(\frac{2}{3}\right)^2 \left(\frac{1}{3}\right)^2 = 6 \times \frac{4}{9} \times \frac{1}{9} = \frac{24}{81} \] Total Probability: \(\frac{1}{81} + \frac{8}{81} + \frac{24}{81} = \frac{33}{81} = \frac{33}{3^4}\).
Step 4: Final Answer:
The probability is \(\frac{33}{3^4}\).