Question

There are 15 students. 5 of them are boys and 10 of them are girls. If 5 students are chosen, what is the probability that there are at least 2 boys?

Answer 1

Reqd. Prob.`=P(A)=567/1001`.

Explanation:

let `A` be the event that, in the selection of `5` students, at least `2` Boys are there.

Then, this event `A` can happen in the following `4` mutually exclusive cases :=

Case (1) :

Exactly `2` Boys out of `5` and `3` Girls ( =5students - 2 boys) out of `10` are selected. This can be done in `(""_5C_2)(""_10C_3)=(5*4)/(1*2)*(10*9*8)/(1*2*3)=1200` ways.

Case (2) :=

Exactly `3B` out of `5B` & `2G` out of `10G`.
No. of ways`=(""_5C_3)(""_10C_2)=10*45=450`.

Case (3) :=

Exactly `4B` & `1G`, no. of ways`=(""_5C_4)(""_10C_1)=50`.

Case (4) :=

Exactly `5B` & `0G` (no G), no. of ways`=(""_5C_5)(""_10C_0)=1`.

Therefore, total no. of outcomes favourable to the occurrence of the event `A=1200+450+50+1=1701`.

Finally, `5` students out of `15` can be selected in `""_15C_5=(15*14*13*12*11)/(1*2*3*4*5)=3003` ways., which is the total no. of outcomes.

Hence, the Reqd. Prob.`=P(A)=1701/3003=567/1001`.

Enjoy Maths.!

Answer 2

Probability of at least 2 boys = P[(2 boys & 3 girls) + (3 boys & 2 girls) + (4 boys & 1 girl) + (5 boys & 0 girl)]`=0.5663`

Explanation:

`p_(2 boys &3 girls) = (C(5,2)xx(C(10,3)))/((C(15,5))`
`=(10xx120)/3003=1200/3003=0.3996`

`p_(3 boys &2 girls) = (C(5,3)xx(C(10,2)))/((C(15,5))`
`=(10xx45)/3003=450/3003=0.1498`

`p_(4 boys &1 girl) = (C(5,4)xx(C(10,1)))/((C(15,5))`
`=(5xx10)/3003=50/3003=0.0166`

`p_(5 boys &0 girl) = (C(5,5)xx(C(10,0)))/((C(15,5))`
`=(1xx1)/3003=1/3003=0.0003`

Probability of at least 2 boys = P[(2 boys & 3 girls) + (3 boys & 2 girls) + (4 boys & 1 girl) + (5 boys & 0 girl)]

`=0.3996 + 0.1498+0.0166+0.0003=0.5663`