JEE Main 2015 · 11 Apr · Shift 1 · Mathematics Q87

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MathematicsSingle correct+4 / -1Question 87 of 90

If the mean and the variance of a binomial variate X are 2 and 1 respectively, then the probability that X takes a value greater than or equal to one is :

\dfrac1{16}

\dfrac9{16}

\dfrac34

\dfrac{15}{16}

Answer and worked solution

Official answer: D

$np=2$ and $npq=1$ give $q=\frac12$ and $n=4$ . $P(X\ge1)=1-\left(\frac12\right)^4=\frac{15}{16}$ .

Topics:Binomial distribution

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JEE Main 2015 · 11 Apr · Shift 1, Mathematics question 87