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.

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.

cdquestions Admin · Accepted Answer

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}$ .

Answer

$\frac{27}{119}$

Answer

$\frac{9}{238}$

Answer

$\frac{18}{119}$

Answer

$\frac{54}{119}$

Answer

None of these

List-I	List-II (Adverbs)
(A) P(exactly 2 heads)	(I) \(\frac{1}{4}\)
(B) P(at least 1 head)	(II) \(1\)
(C) P(at most 2 heads)	(III) \(\frac{3}{4}\)
(D) P(exactly 1 head)	(IV) \(\frac{1}{2}\)

LIST-I(EVENT)	LIST-II(PROBABILITY)
(A) The sum of the number is greater than 11	(i) 0
(B) The sum of the number is 4 or less	(ii) 1/15
(C) The sum of the number is 4	(iii) 2/15
(D) The sum of the number is 4	(iv) 3/15

The Correct Option is A

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