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)=5671001.

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 (5C2)(10C3)=54121098123=1200 ways.

Case (2) :=

Exactly 3B out of 5B & 2G out of 10G.
No. of ways=(5C3)(10C2)=1045=450.

Case (3) :=

Exactly 4B & 1G, no. of ways=(5C4)(10C1)=50.

Case (4) :=

Exactly 5B & 0G (no G), no. of ways=(5C5)(10C0)=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 15C5=151413121112345=3003 ways., which is the total no. of outcomes.

Hence, the Reqd. Prob.=P(A)=17013003=5671001.

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:

p2boys&3girls=C(5,2)×(C(10,3))(C(15,5))
=10×1203003=12003003=0.3996

p3boys&2girls=C(5,3)×(C(10,2))(C(15,5))
=10×453003=4503003=0.1498

p4boys&1girl=C(5,4)×(C(10,1))(C(15,5))
=5×103003=503003=0.0166

p5boys&0girl=C(5,5)×(C(10,0))(C(15,5))
=1×13003=13003=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