Somebody please solve this
A bag contains 5 red balls 4 blue balls and m green balls.Two balls are drawn at random from the bag.If probability of both being green is 1/7,then find m.
so there are m+9 balls
prob(green,green) = (m/(m+9) (m-1)/(m+8) = 1/7
(m+9)(m+8) = 7m(m-1)
7m^2 - 7m = m^2 + 17m + 72
6m^2 - 24m - 72 = 0
we know m has to be a whole number, so this MUST factor, and sure enough
(m-6)(6m + 12) = 0
so m = 6 , or m = -2, the last is no good
so m=6
Thanks Nikhat
To solve this problem, we'll use the concept of probability and combinatorics.
Let's break down the problem step by step:
1. We know that the bag contains 5 red balls, 4 blue balls, and m green balls.
2. We are drawing two balls at random from the bag, which means we have to consider all possible combinations of two balls.
3. For the probability of both balls being green to be 1/7, we need to find the value of m.
To find the probability of drawing two green balls, we need to consider the total number of ways we can select 2 balls from the bag and divide it by the total number of favorable outcomes (in this case, both balls being green).
Let's calculate the total number of ways we can select 2 balls from the bag:
Total ways = (Total number of balls in the bag)C2
= (5 + 4 + m)C2
Now, let's calculate the favorable outcomes (both balls being green):
Favorable outcomes = (Number of green balls)C2
= mC2
Given that the probability of both balls being green is 1/7, we can set up the equation:
Favorable outcomes / Total ways = 1/7
(mC2) / ((5 + 4 + m)C2) = 1/7
We can simplify this equation further:
(m! / (2!(m - 2)!)) / ((9 + m)! / (2!(7 + m)!)) = 1/7
(m! / (2!(m - 2)!)) * ((2!(7 + m)!) / (9 + m)!) = 1/7
(m(m - 1)) / ((9 + m)(8 + m)) = 1/7
7m(m - 1) = (9 + m)(8 + m)
7m^2 - 7m = 72 + 17m + m^2
6m^2 - 24m - 72 = 0
Now, we can solve this quadratic equation to find the value of m.