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MHTCET 5 May 2025 Shift 2 Paper

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Question : 149 of 150

Marks: +1, -0

If

\mathrm{X} \sim B(35, \mathrm{p})

7 \mathrm{P}(\mathrm{X}=0)=\mathrm{P}(\mathrm{X}=1)

then the value of

\frac{\mathrm{P}(\mathrm{X}=15)}{\mathrm{P}(\mathrm{X}=20)}

is equal to

$\frac{3125}{7776}$
3125
7776
$\frac{625}{1296}$

Solution:

Concept:

Use the binomial probability formula

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

, where

q = 1-p

.

Explanation:

Given

7P(X=0) = P(X=1)

and

X \sim B(35, p)

.

P(X=0) = q^{35}

and

P(X=1) = 35pq^{34}

.

So,

7q^{35} = 35pq^{34}

.

Dividing by

q^{34}

gives

7q = 35p

, so

q = 5p

.

Since

p + q = 1

, we get

p = \frac{1}{6}

and

q = \frac{5}{6}

.

Now,

\frac{P(X=15)}{P(X=20)} = \frac{{}^{35}C_{15}p^{15}q^{20}}{{}^{35}C_{20}p^{20}q^{15}}

.

Since

{}^{35}C_{15} = {}^{35}C_{20}

, this simplifies to

\left(\frac{q}{p}\right)^{5}

.

Thus,

\frac{P(X=15)}{P(X=20)} = \left(\frac{5/6}{1/6}\right)^{5} = 5^{5} = 3125

.

Answer:

3125

(Option B).

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