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the probability of a shooter hitting a target is f
Question:
The probability of a shooter hitting a target is \( \frac{3}{4} \). Find the probability of hitting the target exactly 4 times in 5 shots.
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For “exactly \(r\)” successes in repeated independent trials, always apply the binomial formula \( {n \choose r}p^r q^{n-r} \).
MHT CET - 2026
MHT CET
Updated On:
May 12, 2026
\(\frac{405}{1024}\)
\(\frac{243}{1024}\)
\(\frac{405}{512}\)
\(\frac{81}{256}\)
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The Correct Option is
A
Solution and Explanation
Concept:
This problem follows the
Binomial Distribution
. If probability of success = \(p\) and probability of failure = \(q\), then probability of exactly \(r\) successes in \(n\) trials is: \[ P(X=r) = {n \choose r} p^r q^{\,n-r} \]
Step 1:
Identify the parameters. \[ n = 5,\quad r = 4 \] \[ p = \frac{3}{4}, \quad q = 1 - p = \frac{1}{4} \]
Step 2:
Apply the binomial probability formula. \[ P(X=4) = {5 \choose 4} \left(\frac{3}{4}\right)^4 \left(\frac{1}{4}\right) \]
Step 3:
Simplify the expression. \[ {5 \choose 4} = 5 \] \[ P(X=4) = 5 \times \frac{81}{256} \times \frac{1}{4} \] \[ = 5 \times \frac{81}{1024} \] \[ = \frac{405}{1024} \] Thus, \[ \boxed{\frac{405}{1024}} \]
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