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MHTCET 23 Apr 2025 Shift 2 Paper

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Question : 114 of 150

Marks: +1, -0

If

X \sim B\left(6, \frac{1}{2}\right)

P(|X-2| \leqslant 1)=

$\frac{31}{32}$
$\frac{41}{64}$
$\frac{51}{64}$
$\frac{63}{64}$

Solution:

Concept:

For a binomial distribution,

P(X=r)={}^{n}C_{r}\,p^{r}q^{,n-r}

.

Explanation:

Given

X \sim B\left(6,\frac{1}{2}\right)

, so

n=6

,

p=\frac{1}{2}

,

q=\frac{1}{2}

.

The condition

|X-2|\leq 1

gives:

-1 \leq X-2 \leq 1 \Rightarrow 1 \leq X \leq 3.

Thus,

P(|X-2|\leq 1)=P(X=1)+P(X=2)+P(X=3).

Using the binomial formula,

P(X=r)={}^{6}C_{r}\left(\frac{1}{2}\right)^{r}\left(\frac{1}{2}\right)^{6-r}={}^{6}C_{r}\left(\frac{1}{2}\right)^{6}.

So,

P(X=1)=\frac{6}{64},\quad P(X=2)=\frac{15}{64},\quad P(X=3)=\frac{20}{64}.

Therefore,

P(|X-2|\leq 1)=\frac{6+15+20}{64}=\frac{41}{64}.

Answer:

\frac{41}{64}

, which is option B.

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