MHT CET · Maths · Probability
If the mean and the variance of Binomial variate \(X\) are 2 and 1 respectively, then the probability that X takes a value greater than or equal to one is
- A \(\frac{1}{16}\)
- B \(\frac{9}{16}\)
- C \(\frac{3}{4}\)
- D \(\frac{15}{16}\)
Answer & Solution
Correct Answer
(D) \(\frac{15}{16}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& \text {Mean }=\mathrm{np}=2 \text { and variance}=\mathrm{npq}=1 \\
& \therefore \mathrm{q}=\frac{1}{2}
\end{aligned}\)
\(\text {Also, } p=1-q=1-\frac{1}{2}=\frac{1}{2} \)
\( \mathrm{n}=4\)
\(\therefore P(X \geq 1) =1-P(X=0) \)
\( =1-{ }^4 C_0 p^0 q^4 \)
\( =1-(1)\left(\frac{1}{2}\right)^0\left(\frac{1}{2}\right)^4 \)
\( =1-\frac{1}{16} \)
\( =\frac{15}{16}\)
& \text {Mean }=\mathrm{np}=2 \text { and variance}=\mathrm{npq}=1 \\
& \therefore \mathrm{q}=\frac{1}{2}
\end{aligned}\)
\(\text {Also, } p=1-q=1-\frac{1}{2}=\frac{1}{2} \)
\( \mathrm{n}=4\)
\(\therefore P(X \geq 1) =1-P(X=0) \)
\( =1-{ }^4 C_0 p^0 q^4 \)
\( =1-(1)\left(\frac{1}{2}\right)^0\left(\frac{1}{2}\right)^4 \)
\( =1-\frac{1}{16} \)
\( =\frac{15}{16}\)
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