JEE Main 2022 · 24 Jun · Shift 1 · Mathematics Q17

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MathematicsSingle correct+4 / -1Question 17 of 90

If a random variable $X$ follows the Binomial distribution $B(33,p)$ such that $3P(X=0)=P(X=1)$ , then the value of $\dfrac{P(X=15)}{P(X=18)}-\dfrac{P(X=16)}{P(X=17)}$ is equal to :

1320

1088

\dfrac{120}{1331}

\dfrac{1088}{1089}

Answer and worked solution

Official answer: A

Find $p$ . $3q^{33}=33pq^{32}$ gives $3q=33p$ , so $p=\frac1{12}$ , $q=\frac{11}{12}$ , $\frac qp=11$ .

Ratios. $\binom{33}{15}=\binom{33}{18}$ , so $\dfrac{P(15)}{P(18)}=\left(\dfrac qp\right)^3=1331$ . Likewise $\dfrac{P(16)}{P(17)}=\dfrac qp=11$ .

Value. $1331-11=1320$ .

(A)

Topics:Binomial distribution Properties of binomial coefficients

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JEE Main 2022 · 24 Jun · Shift 1, Mathematics question 17