bag contains 4 red, 5 yellow and 6 green balls. 3 balls are drawn randomly. What is the probability that the balls drawn contain no red ball ?
Answers
Step-by-step explanation:
Initially, there are (3 + 4 + 5) = 12 balls in the bag.
Now, two types of draw can be done.
(a) Drawing without replacement:
Ways of drawing 1 red ball out of 3 red balls = (3C1) = 3.
Ways of drawing 1 yellow ball out of 5 yellow balls = (5C1) = 5.
Ways of drawing 1 green ball out of 4 green balls = (4C1) = 4.
Ways of drawing 3 balls out of 12 = (12C3) = 220.
Therefore, probability of getting 3 balls of different colours from first three draws without replacement = [(3*5*4) / 220] = (60 / 220) = (3 / 11).
(b) Drawing with replacement:
Ways of drawing 1 red ball out of 3 red balls = (3C1) = 3.
Ways of drawing 1 yellow ball out of 5 yellow balls = (5C1) = 5.
Ways of drawing 1 green ball out of 4 green balls = (4C1) = 4.
Ways of drawing 3 balls out of 12 = (12C1)*(12C1)*(12C1) = (12*12*12) = 1728.
Therefore, probability of getting 3 balls of different colours from first three draws with replacement = [(3*5*4) / 1728] = (60 / 1728) = (5 / 144).