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CBSE Class 12-science Answered

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 

 

begin mathsize 16px style probability space of space taking space 1 st space ball space red space ball equals 4 over 13
probability space of space taking space 2 nd space ball space green space ball equals 6 over 12 space left parenthesis no space replacement right parenthesis
probability space of space taking space 3 rd space ball space green space ball equals 5 over 11 space left parenthesis no space replacement right parenthesis
so space in space total
4 over 13 cross times 6 over 12 cross times 5 over 11 end style

 

Asked by sachinkumarmishra1 | 22 Jan, 2020, 05:50: PM
Expert Answer
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
therefore space R e q u i r e d space p r o b a b i l i t y space equals space fraction numerator C presubscript 1 presuperscript 4 cross times C presubscript 1 presuperscript 6 cross times C presubscript 1 presuperscript 5 over denominator C presubscript 1 presuperscript 13 cross times C presubscript 1 presuperscript 12 cross times C presubscript 1 presuperscript 11 end fraction equals fraction numerator 4 cross times 5 cross times 6 over denominator 13 cross times 12 cross times 11 end fraction
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