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?

2 Answers
Aug 13, 2016

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.!

Aug 13, 2016

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