GATE CSE 2005 | Question: 51

Question

Box $P$ has $2$ red balls and $3$ blue balls and box $Q$ has $3$ red balls and $1$ blue ball. A ball is selected as follows: (i) select a box (ii) choose a ball from the selected box such that each ball in the box is equally likely to be chosen. The probabilities of selecting boxes $P$ and $Q$ are $\dfrac{1}{3}$ and $\dfrac{2}{3}$ respectively. Given that a ball selected in the above process is a red ball, the probability that it came from the box $P$ is:

• A. $\dfrac{4}{19}$

• B. $\dfrac{5}{19}$

• C. $\dfrac{2}{9}$

• D. $\dfrac{19}{30}$

Answer 1

Just straight way use bayes theorem you wl get 4/19

Answer 2

We can solve this question with the help of probability tables $\rightarrow$

 

	$Red$	$Blue$
$P$	$\color{red}{\frac{1}{3} \times \frac{2}{5}}$	$\frac{1}{3}\times\frac{3}{5}$	$\frac{1}{3}$
$Q$	$\frac{2}{3} \times \frac{3}{4}$	$\frac{2}{3}\times\frac{1}{4}$	$\frac{2}{3}$
	$\color{green}{\frac{19}{30}}$	$\frac{11}{30}$	$1$

 

Given that a ball selected in the above process is a red ball, the probability that it came from the box $P$ $ = \frac{\color{red}{\frac{1}{3} \times \frac{2}{5}}}{\color{green}{\frac{19}{30}}} = \frac{4}{19}$

Answer 3

BY Bayes theorem we get this

 

Answer 4

Box P has 2 red balls and 3 blue balls and Q has 3 red balls and 1 blue ball.

Let p:selection of box P ,q: selection of box Q ,r: Red ball being selected

P(p)=1/3  P(q)=2/3

P(r)=P(r|p)*P(p)+P(r|q)*P(q)=(2/5)*(1/3)+(3/4)*(2/3)=38/60

P(p|r)=P(p∩r)/P(r)=(P(r|p)*P(p))/(38/60)=(2/15)*(60/38)=4/19