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Probability Part 1

Section: Mathematics

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Question : 24 of 100

Marks: +1, -0

An ordinary dice is rolled for a certain number of times. If the probability of getting an odd number 2 times is equal to the probability of getting an even number 3 times, then the probability of getting an odd number for odd number of times is

[24 Feb 2021 Shift 1]

$\;\frac{1}{32}$
$\;\frac{3}{16}$
$\;\frac{5}{16}$
$\;\frac{1}{2}$

Solution:

Given,

P

(odd number 2 times)

=P(

even number 3 times)

\Rightarrow\;\; {}^{n}C_{2}\left(\;\frac{1}{2}\right)^{n}={}^{n}C_{3}\left(\;\frac{1}{2}\right)^{n}

where,

n=

Number of times the die is thrown.

\Rightarrow\;\; \;\frac{n!}{2!(n-2)!} \times \left(\;\frac{1}{2}\right)^{n} = \;\frac{n!}{3!(n-3)!} \times \left(\;\frac{1}{2}\right)^{n}

\Rightarrow\;\; \;\frac{3! \times (n-3)!}{2! \times (n-2)!}=1

\Rightarrow\;\; \;\frac{3 \times 2! \times (n-3)!}{2! \times (n-2)(n-3)!}=1 \Rightarrow 3=(n-2)

\Rightarrow\;\; n=5

\therefore

Probability of getting an odd number, odd number of times

\;\; =P(1)+P(3)+P(5)

\;\; ={}^{5}C_{1}\left(\;\frac{1}{2}\right)^{5}+{}^{5}C_{3}\left(\;\frac{1}{2}\right)^{5}+{}^{5}C_{5}\left(\;\frac{1}{2}\right)^{5}

\;\; =\;\frac{16}{2^{5}}=\;\frac{16}{32}=\;\frac{1}{2}

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