Question:medium

Four patients are treated with an intervention that is successful 90% of the time. What is the probability of two successes?

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For binomial problems, carefully apply $P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$. Always recheck calculations to avoid mismatches.
Updated On: Feb 19, 2026
  • 0.0324
  • 0.486
  • 0.324
  • 0.0486
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: Identify Distribution. This is a binomial probability scenario. We have a fixed number of independent trials (4 patients), each with two possible outcomes (success or failure), and a constant probability of success ($p = 0.9$).
Step 2: State Formula. The formula for the probability of exactly $k$ successes in $n$ trials is:
\[P(X = k) = \binom{n}{k} p^{k} (1-p)^{n-k}\]
Step 3: Input Values. We have $n = 4$, $k = 2$, $p = 0.9$, and $(1-p) = 0.1$. Substituting these values gives:
\[P(X = 2) = \binom{4}{2} (0.9)^2 (0.1)^2\]
Step 4: Calculate. First, calculate the binomial coefficient:
\[\binom{4}{2} = \frac{4!}{2! \cdot 2!} = 6\] Then, calculate the probability:
\[P(X = 2) = 6 \times (0.81) \times (0.01) = 6 \times 0.0081 = 0.0486\]
Step 5: Select Option. The calculated probability is 0.0486, which corresponds to option (D).

% Correction Note
Note: If the official key indicates 0.324, this value represents the probability of 3 successes, not 2. Please verify the key.
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