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
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
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This is a binomial probability problem with four trials; add the probabilities of 0, 1 and 2 survivals.
Updated On: Nov 13, 2026
- \(\frac{31}{3^4}\)
- \(\frac{32}{3^4}\)
- \(\frac{33}{3^4}\)
- \(\frac{35}{3^4}\)
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept
Number of survivors follows a binomial distribution with \(n = 4\), \(p = \dfrac{2}{3}\), \(q = \dfrac{1}{3}\).
Step 2: Individual Probabilities
\(P(0) = \left(\dfrac{1}{3}\right)^4 = \dfrac{1}{81}\). \(P(1) = 4\cdot\dfrac{2}{3}\cdot\left(\dfrac{1}{3}\right)^3 = \dfrac{8}{81}\). \(P(2) = 6\cdot\left(\dfrac{2}{3}\right)^2\left(\dfrac{1}{3}\right)^2 = \dfrac{24}{81}\).
Step 3: Add Them
\(P(X \le 2) = \dfrac{1 + 8 + 24}{81} = \dfrac{33}{81} = \dfrac{33}{3^4}\).
Step 4: Compare the options
The probability is \(\dfrac{33}{3^4}\), the third option.
Final Answer:
The probability that at most two components survive is 33 over 3 to the power 4. \[ \boxed{\text{(C) }\dfrac{33}{3^4}} \]
Number of survivors follows a binomial distribution with \(n = 4\), \(p = \dfrac{2}{3}\), \(q = \dfrac{1}{3}\).
Step 2: Individual Probabilities
\(P(0) = \left(\dfrac{1}{3}\right)^4 = \dfrac{1}{81}\). \(P(1) = 4\cdot\dfrac{2}{3}\cdot\left(\dfrac{1}{3}\right)^3 = \dfrac{8}{81}\). \(P(2) = 6\cdot\left(\dfrac{2}{3}\right)^2\left(\dfrac{1}{3}\right)^2 = \dfrac{24}{81}\).
Step 3: Add Them
\(P(X \le 2) = \dfrac{1 + 8 + 24}{81} = \dfrac{33}{81} = \dfrac{33}{3^4}\).
Step 4: Compare the options
The probability is \(\dfrac{33}{3^4}\), the third option.
Final Answer:
The probability that at most two components survive is 33 over 3 to the power 4. \[ \boxed{\text{(C) }\dfrac{33}{3^4}} \]
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