Question

A bag contains some red, some green and 9 blue balls. Ratio of number of red to green ball in the bag is 5:4. If the probability of drawing a red ball is $\frac{1}{6}$ more than that of drawing a blue ball then find the probability of getting three different coloured ball when three balls are drawn from the bag one after another and without replacement.

• A. $\frac{27}{119}$
• B. $\frac{9}{238}$
• C. $\frac{18}{119}$
• D. $\frac{54}{119}$
• E. None of these

Answer 1

Answer: (A)

$\frac{27}{119}$

Answer 2

Let, number of red and green balls in the bag be ‘5x’ and ‘4x’, respectively.
Total number of balls in the bag = 5x + 4x + 9 = 9x + 9
So, $\frac{(5x – 9)}{(9x + 9) }=\frac{ 1}{6}$
Or, 30x – 54 = 9x + 9
Or, 21x=63
Or, x=3
Number of red balls = 5 × 3 = 15
Number of green balls = 4 × 3 = 12
Total number of balls = 15 + 12 + 9 = 36
Desired probability = $\frac{{^{15}C1x^{12}C_1x^9C_1}/^{36}}{C_3}$ = $\frac{(15 × 12 × 9 × 6)}{ (36 × 35× 34) }= \frac{27}{119}$
So, The correct option is (A) : $\frac{27}{119}$.