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a bag contains 4 red ,6 green, 3 white balls ,three balls are drawn one after one the other without replacement,find the probability that the balls are red,green and green  My doubt is --How can we solve this question by combination ? The explanation given is
Asked by sachinkumarmishra1 | 22 Jan, 2020, 05:50: PM
Total number of balls = 13
Three balls are drawn one after the other without replacement
First ball can be drawn in 13C1 no. of ways
No. of balls left = 12
Second ball can be drawn in 12C2 no. of ways
Third ball can be drawn in 11C1 no. of ways
So, the total number of ways = 13C1 x 12C1 x 11C1
First ball should be red, so the number of ways of drawing red ball = 4C1
Second ball should be green, so the no. of ways of drawing 1st green ball = 6C1
Only 5 green balls are left, so the no. of ways of drawing 2nd green ball = 5C1
Answered by Renu Varma | 23 Jan, 2020, 11:04: AM

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