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Enter an inline fraction (a/b) in simplest form. In a class of 10 students there are 4 boys and 6 girls. If a committee of 5
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Answer #1

Here order is not matter so we can use combinations

there are total 10 students and we have to choose 5 students from total 10 students.

So there are 10C5 total ways of selcting committe of 5

= 10c5 = 10!/5!*5! = 10*9*8*7*6*5! /5*4*3*2*1 *5! = 30240/120 = 252

Also there are 4 boys in class of 10 and in commeittee of 5 we need 3 boys

So there are 4C3 total ways of selcting 3 boys in committe of 5

=4C3 = 4!/3!*1! = 4*3*2*1/3*2*1*1 =24/6= 4

Also there are 6 girls in class of 10 and in commeittee of 5 we need 2 girls

So there are 6C2 total ways of selcting 2 girls in committe of 5

=6C2 = 6!/2!*4! = 6*5*4*3*2*1/2*1*4*3*2*1 =720/48 = 15

So probability of selecting 3 boys and 2 girls is

4C3 * 6C2 / 10C5 = 4*15/ 252 = 60/252 = 5/21 = 0.2381

So the answer is 5/21

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