Question:
If $X\sim B(n,p)$ is a binomial variate, $8p^2+15p-2=0$ and the product of the mean and variance of $X$ is $\dfrac78$, then $P(X=4)=$
If $X\sim B(n,p)$ is a binomial variate, $8p^2+15p-2=0$ and the product of the mean and variance of $X$ is $\dfrac78$, then $P(X=4)=$
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For a binomial variable, remember: Mean $=np$ and Variance $=npq$.
Updated On: Jun 3, 2026
- ${}^{6}C_{4}\dfrac{5^2}{6^6}$
- ${}^{7}C_{4}\dfrac{4^3}{5^7}$
- ${}^{5}C_{4}\dfrac{3}{4^5}$
- ${}^{8}C_{4}\dfrac{7^4}{8^8}$
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The Correct Option is D
Solution and Explanation
Step 1: Concept
Use the binomial mean and variance formulas.
Step 2: Meaning
Since \[ 8p^2+15p-2=0, \] the valid probability root is \[ p=\frac18. \] Hence \[ q=1-p=\frac78. \]
Step 3: Analysis
Mean and variance are \[ np,\qquad npq. \] Given \[ (np)(npq)=\frac78. \] Thus \[ n^2\left(\frac18\right)\left(\frac{7}{64}\right) = \frac78. \] Therefore \[ n^2=64 \Rightarrow n=8. \] Hence \[ P(X=4) = {}^{8}C_{4} \left(\frac18\right)^4 \left(\frac78\right)^4. \] \[ = {}^{8}C_{4}\frac{7^4}{8^8}. \]
Step 4: Conclusion
Therefore the required probability is \[ {}^{8}C_{4}\frac{7^4}{8^8}. \]
Final Answer: (D)
Use the binomial mean and variance formulas.
Step 2: Meaning
Since \[ 8p^2+15p-2=0, \] the valid probability root is \[ p=\frac18. \] Hence \[ q=1-p=\frac78. \]
Step 3: Analysis
Mean and variance are \[ np,\qquad npq. \] Given \[ (np)(npq)=\frac78. \] Thus \[ n^2\left(\frac18\right)\left(\frac{7}{64}\right) = \frac78. \] Therefore \[ n^2=64 \Rightarrow n=8. \] Hence \[ P(X=4) = {}^{8}C_{4} \left(\frac18\right)^4 \left(\frac78\right)^4. \] \[ = {}^{8}C_{4}\frac{7^4}{8^8}. \]
Step 4: Conclusion
Therefore the required probability is \[ {}^{8}C_{4}\frac{7^4}{8^8}. \]
Final Answer: (D)
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