Question:
If in \(6\) trials, X is a binomial random variable which follows the relation \(9P(x = 4) = P(x = 2)\), then the probability of failure is...
If in \(6\) trials, X is a binomial random variable which follows the relation \(9P(x = 4) = P(x = 2)\), then the probability of failure is...
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Write both probabilities with the binomial formula and solve for p/q.
Updated On: Oct 1, 2026
- \(0.125\)
- \(0.25\)
- \(0.375\)
- \(0.75\)
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The Correct Option is D
Solution and Explanation
Step 1: Binomial Formula:
\(P(X=r)=\binom6r p^rq^{6-r}\) with \(q=1-p\).
Step 2: Form the Equation:
\[ 9\binom64p^4q^2=\binom62p^2q^4 \]
Both binomials equal 15, so they cancel: \(9p^4q^2=p^2q^4\).
Step 3: Simplify:
Divide by \(p^2q^2\): \(9p^2=q^2\), so \(q=3p\) (probabilities are positive).
Step 4: Solve:
\(p+q=1\Rightarrow p+3p=1\Rightarrow p=\tfrac14\). The probability of failure is \(q=\tfrac34=0.75\). Option (D).
Final Answer:
The probability of failure is 0.75, option (D). \[ \boxed{\text{(D) } 0.75} \]
\(P(X=r)=\binom6r p^rq^{6-r}\) with \(q=1-p\).
Step 2: Form the Equation:
\[ 9\binom64p^4q^2=\binom62p^2q^4 \]
Both binomials equal 15, so they cancel: \(9p^4q^2=p^2q^4\).
Step 3: Simplify:
Divide by \(p^2q^2\): \(9p^2=q^2\), so \(q=3p\) (probabilities are positive).
Step 4: Solve:
\(p+q=1\Rightarrow p+3p=1\Rightarrow p=\tfrac14\). The probability of failure is \(q=\tfrac34=0.75\). Option (D).
Final Answer:
The probability of failure is 0.75, option (D). \[ \boxed{\text{(D) } 0.75} \]
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