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?

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.!

#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#