Study the following information carefully and answer the given questions.
There were some red, some green and some white balls in a bag. Probability of drawing 1 white ball from the bag is . `(1)/(3)` If there were 6 more red balls in the bag, the probability of getting 1 green ball would have been `(1)/(30)` less.
If sum of red, green and white balls is 30, then what is the probability of drawing 2 white ballsfrom the bag?

Options

A. `(151)/(870)`
B. `(151)/(435)`
C. `(31)/(145)`
D. `(141)/(435)`

Answer

B

Explanation

Let the number of white balls = w, the number of green balls = g and the number of red balls = r, then g + w = 16
Given `(w)/((w + g + r)) = (1)/(3)`
`rArr w = ((16 + r))/(3)`
`rArr g = 16 - w = 16 - ((16 + r))/(3) = ((32 - r))/(3)`
According to questions
`rArr (g)/(( w + g + r)) - (g)/((w + g + r + 6)) = (1)/(30)`
`rArr ( 2 ( 32 - r))/((w + g + r) ( w + g + r + 6)) = (1)/(30)`
`rArr ( 2 ( 32 - r))/(( 16 + r) ( 22 + r)) = (1)/(30)`
`rArr r^(2) + 98 r + 2272 = 0 `
`rArr ^(2) + 112 r - 14 r - 2272 = 0` So, ` g = (( 32 - r))/( 3) = ((32 - 14))/(3) = 6 and w = 16 - g =10 `
Hence, the required probability `= (14_(C_(2)))/(20_(C_(2))) + (6_(C_(2)))/(30_(C_(2)))+(10_(C_(2)))/(30_(C_(2)))=(14 xx 13)/(30 xx 29) + (6 xx 5)/(20 xx 29)+ (10 xx 9)/(30 xx 29) = (182 + 30 + 90)/(30 xx 29) = (151)/(435)`