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BITSAT 2017 Paper

Section: Mathematics

Question : 120 of 150

Marks: +1, -0

Solution:

Let

A_i(i=2,3,4,5)

be the event that urn contains 2, 3, 4, 5 white balls and let B be the event that two white balls have been drawn then we have to find

P\left(\frac{A_5}{B}\right)

.Since the four events are equally likely we have

P(A_2)=P(A_3)=P(A_4)=P(A_5)=\frac{1}{4}

P\left(\frac{B}{A_2}\right)

is probability of event that the urn contains 2 white balls and both have been drawn.

\therefore P\left(\frac{B}{A_2}\right)=\frac{{}^{2}C_{2}}{{}^{5}C_{2}}=\frac{1}{10}

Similarly

P\left(\frac{B}{A_3}\right)=\frac{{}^{3}C_{2}}{{}^{5}C_{2}}=\frac{3}{10}

P\left(\frac{B}{A_4}\right)=\frac{{}^{4}C_{2}}{{}^{5}C_{2}}=\frac{3}{5},\quad P\left(\frac{B}{A_5}\right)=\frac{{}^{5}C_{2}}{{}^{5}C_{2}}=1

By Baye’s theorem,

P\left(\frac{A_5}{B}\right)=\frac{P(A_5)P\left(\frac{B}{A_5}\right)}{P(A_2)P\left(\frac{B}{A_2}\right)+P(A_3)P\left(\frac{B}{A_3}\right)+P(A_4)\left(\frac{B}{A_4}\right)+P(A_5)P\left(\frac{B}{A_5}\right)}

=\frac{\frac{1}{4}\cdot 1}{\frac{1}{4}\left[\frac{1}{10}+\frac{3}{10}+\frac{3}{5}+1\right]}=\frac{10}{20}=\frac{1}{2}

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