Problem Solving

From a group of 3 boys and 3 girls, 4 children are to be randomly selected. What is the probability that equal numbers of boys and girls will be selected?
ANS[spoiler] 4/9[/spoiler]

According to me the answer should be 9/15 = 3/5

4/9 looks wrong to me. Reason being, he is simple taking 2/3 * 2/3 = 4/9

Doesnt make sense...



Here is how I get 3/5 -
Denominator - Total no of ways to get 4 children = 6C4
Numerator - Total no of ways to select 2 girls of 3 * Total no of ways to select 2 boys of 3
=(3C2 * 3C2)/6C4
=9/15 = 3/5

arorag, could you please check again. I doubt the OA. Whats the source of the quest..

Anybody has any other explainations??

Ian Stewart wrote:Yes, the OA (4/9) can't be right here. The answer is 3/5; I used the same approach as in the above post.

Thanks for saying something, punit, Ian! I was concerned too when I saw the answer. Phew.

Hi,

For some reason I am having a hard time understanding the probablity questions that are usually posted. Can one of you solve this in a little more detail (dumbed down).

Thanks for any help

I'll give this a shot. You have three guys and three girls to select from. You have 4 spots. The order in these 4 spots is insignificant. So you have to use the formula of combinations to determine the total number of selection possibilities. This boils down to 6C4 (i.e. six people, 4 spots, and order does not matter). This gives you a total of 15 possibilities.

Now, think of the number of ways in which the number of girls will NOT be equal to the number of guys. Two scenarios will results in this.

Scenario 1: 3 girls are selected and one boy is selected

The number of selections that would fit this scenario = 3

because one of three boys can be selected as the "one boy with the three girls".

Scenario 2: 3 boys are selected and one girl is selected

The number of selections that would fit this scenario = 3

because one of three girls can be selected as the "one girl with the three boys".

Hence, the total number of selections in which the number of boys is not equal to the number of girls is 6.

Hence, the number of selections in which the number of boys IS equal to the nunber of girls = 15 (i.e. total number of selection possibilities) - 6

= 9

Hence, the probability of this equal situation = 9/15

I hope this makes sense.

Thanks, that made a lot of sense. My biggest problem was that I could not figure out how you guys were getting 15. After you mentioned the "formula of combinations", I googled it and found the equation. The problem was much easier after that.

ok, isnt 6C4 equal to 30? 6*5*4*3*2*1/4*3*2*1?

3B 3G --> 4 to be selected

(15 - 3B3*3G1 - 3G3*3B1) = 15 - 1*3 - 1*3 = 9

6c4 = 6x5/2 = 15

Probability = 9/15 = 3/5

Out of 3 girls, choose 2: 3C2= 3

Out of 3 boys, choose 2: 3C2= 3

Therefore, total ways= 3*3 =9

Total ways of choosing 4 persons out of 6= 6C4= 15

Therefore, total probability= 9/15= 3/5

Another brute force way is:
We can have a group of 4 either 1B3G, or 2B2G, or 3B1G.
Case1: 3B1G = B1B2B3G1 or B1B2B3G2 OR B1B2B3G3 = total 3 ways
Case3: 3G1B = similar to above = total 3 ways
Case2: 2B2G = {B1B2G1G2, B1B2G2G3, B1B2G1G3} now replace - B1B2 with B2B3 and afterwards with B3B1 (with same G* combinations).
Total combinations will be 3+3+3 = 9

Case 2 is favourable combination: probability = 9/(3+3+9)=9/15=3/5.
As mentioned this is brute force and cannot come in handy with complex combinations.

please break down your calculation of 6C4 and how that equals 15

please break down your calculation of 6C4 and how that equals 15

stephny wrote:please break down your calculation of 6C4 and how that equals 15

Combination formula : nCr ==> n!/r!(n-r)!

So, 6C4 ==> 6!/4!(6-4)! ==> 6!/4!2! ==> 6x5x4!/4!/2! ==> 6x5/2x1 ==> 3x5 ==> 15

I hope it's clear now!