Question

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

• A. 0.0324
• B. 0.486
• C. 0.324
• D. 0.0486

Hint:

For binomial problems, carefully apply $P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$. Always recheck calculations to avoid mismatches.

Answer 1

The Correct Option is C

Step 1: Recognize distribution type.
This is a binomial probability problem because there are a fixed number of independent trials (4 patients), each with two outcomes (success or failure), and a constant probability of success ($p = 0.9$).
Step 2: Write binomial probability formula.
The probability of exactly $k$ successes in $n$ trials is:
\[ P(X = k) = \binom{n}{k} p^{k} (1-p)^{n-k} \] Step 3: Substitute values.
Here, $n = 4$, $k = 2$, $p = 0.9$, and $(1-p) = 0.1$.
\[ P(X = 2) = \binom{4}{2} (0.9)^2 (0.1)^2 \] Step 4: Simplify calculation.
\[ \binom{4}{2} = \frac{4!}{2! \cdot 2!} = 6 \] \[ P(X = 2) = 6 \times (0.81) \times (0.01) = 6 \times 0.0081 = 0.0486 \] Step 5: Correct option.
The correct probability is 0.0486, which matches option (D).
% Correction Note Note: If the exam’s official key lists 0.324, that corresponds to the probability of 3 successes, not 2. Please double-check the key.