Probability Ques 23
Question-
In bag A there are 5 red balls, $x$ green balls and 7 yellow balls. Probability of drawing one green ball from bag A is $\frac{3}{5}$. In bag B there are $(x-3)$ red balls, $(x-4)$ green balls and 6 yellow balls. 2 balls are drawn from bag $B$. Find the probability that both the balls are of red colour?
(1) $\frac{3}{17}$
(2) $\frac{1}{7}$
(3) $\frac{5}{17}$
(4) $\frac{2}{17}$
(5) None of these
(IBPS RRBs Officer CWE (Main Exam) 05.11.2017)
Show Answer
Correct Answer: (1)
Solution: (1)
For bag A,
Total number of balls $=5+x+7=12+x$
Possible outcomes $=x$
$\therefore$ Required probability of drawing one green ball $=\frac{x}{12+x}=\frac{3}{5}$
$\Rightarrow 5 x=36+3 x$
$\Rightarrow 5 x-3 x=36$
$\Rightarrow 2 x=36 \Rightarrow x=18$
$\therefore$ Number of balls in bag $B$ $=x-3+x-4+6$ $=2 x-1$ $=2 \times 18-1=35$
Number of red balls $=x-3$ $=18-3=15$
$\therefore$ Probability of drawing two red balls $=\frac{{ }^{15} C_{2}}{{ }^{35} C_{2}}$ $=\frac{15 \times 14}{35 \times 34}=\frac{3}{17}$
BETA
