A bag contains 5 red, 4 blue and 'm' green balls. If probability of getting green balls, when 2 balls are selected at random is 117, find number of balls the bag. Ans : 15
Answers
Given:
(i) A bag contains 5 red, 4 blue and 'm' green balls.
(ii) The probability of getting green balls, when 2 balls are selected at random is 1/7
To find:
(i) The number of balls present in the bag.
Solution:
Total number of balls present in bag = (5+4+m)
= 9+m
Probability of 'm' green balls in a total of 9+m balls is
= m/(9+m)
When one green ball is selected,
total green balls = m-1
total balls = 9+m-1 = 8+m
So, probability of 'm-1' green balls in '8+m' balls is
= (m-1)/(8+m)
So, [m/(9+m)]*[(m-1)/(8+m)] = 1/7
⇒ 7m² - 7m = 72 + 17m + m²
⇒ 6m² - 24m - 72 = 0
⇒ m² - 4m - 12 = 0
⇒ m² -6m + 2m - 12 = 0
⇒ m(m-6) + 2(m-6) = 0
⇒ (m-6)(m+2) = 0
By Zero Product Rule,
m = 6 [-2 rejected as number of balls cant be negative]
So, there are 6 green balls.
Total balls present = 5+4+6 = 15
15 balls are present.