Question
A bag contains 5 green balls, 3 black balls, 4 red balls
and 6 blue balls. Ram bets that he will pick two balls at random without replacement and both will be of the same colour. Shyam takes out all the red balls from the bag and then bets that when he picks 2 balls, at random and without replacement, from the bag, both will be of different colours. Find the difference in probability between Ram and Shyam winning the bet.Solution
Total number of balls in the bag initially = 5 + 3 + 4 + 6 = 18 Probability of picking 2 green balls at random = 5C2 ÷ 18C2 = (10/153) Probability of picking 2 black balls at random = 3C2 ÷ 18C2 = (3/153) Probability of picking 2 red balls at random = 4C2 ÷ 18C2 = (6/153) Probability of picking 2 blue balls at random = 6C2 ÷ 18C2 = (15/153) So, probability of Ram winning the bet = (10 + 3 + 6 + 15) ÷ 153 = (34/153) = (2/9) After picking out all the red balls, total number of balls left in the bag = 5 + 3 + 6 = 14 Probability of Shyam picking two green balls = 5C2 ÷ 14C2 = (10/91) Probability of Shyam picking two black balls = 3C2 ÷ 14C2 = (3/91) Probability of Shyam picking two blue balls = 6C2 ÷ 14C2 = (15/91) So, probability of Shyam picking two balls of the same colour = (10 + 3 + 15) ÷ 91 = (28/91) = (4/13) So, probability of Shyam picking two balls of different colours i.e. probability of Shyam winning the bet = 1 - (4/13) = (9/13) So, difference in probability = (9/13) - (2/9) = (81/117) - (26/117) = (55/117)
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