In a bag, there are four different colour of balls namely Red, Blue, Green and Yellow. The ratio between the number of blue and yellow balls in the bag is 3:5 respectively. The number of green balls in the bag is 2.4 times the number of blue balls in the bag. The average of the number of Red, Blue, Green and Yellow balls in the bag is 25. If the probability of picking two Red balls from the bag is (46/825), then find out the probability of picking two balls from the bag such that none of them is Red and Yellow.

Solution

The ratio between the number of blue and yellow balls in the bag is 3:5 respectively. Let’s assume the number of blue and yellow balls in the bag is ‘3a’ and ‘5a’ respectively. The number of green balls in the bag is 2.4 times the number of blue balls in the bag. number of green balls in the bag = 2.4x3a = 7.2a The average of the number of Red, Blue, Green and Yellow balls in the bag is 25. number of Red, Blue, Green and Yellow balls in the bag = 25x4 = 100 number of Red balls in the bag = 100-(3a+5a+7.2a) = (100-15.2a) If the probability of picking two red balls from the bag is (46/825). [(100-15.2a)x(99-15.2a)]/(100x99) = (46/825) [(100-15.2a)x(99-15.2a)]/(4x3) = 46 [(100-15.2a)x(99-15.2a)] = 46x12 [(100-15.2a)x(99-15.2a)] = 24x23 So (100-15.2a) = 24 15.2a = 100-24 a = 5 Similarly (99-15.2a) = 23 a = 5 Number of Blue and Green balls in the bag = (7.2a+3a) = 10.2a = 10.2x5 = 51 Probability of picking two balls from the bag such that none of them is Red and Yellow = (51x50)/(100x99) = (17)/(2x33) = 17/66