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?

Question

Isaiah Anthony · Accepted Answer

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

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.!

Nathan Axel · Answer

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#

#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#

Kennedy Adams · Answer

undefined To find the probability that there are at least 2 boys when 5 students are chosen from a group of 15 (5 boys and 10 girls), we need to calculate the probability of selecting 2 boys, 3 boys, 4 boys, or 5 boys, and then add these probabilities together.

The total number of ways to choose 5 students out of 15 is given by the combination formula:

$$ \binom{15}{5} = \frac{15!}{5!(15-5)!} = 3003 $$

The number of ways to choose exactly 2 boys and 3 girls is:

$$ \binom{5}{2} \times \binom{10}{3} = \frac{5!}{2!(5-2)!} \times \frac{10!}{3!(10-3)!} = 10 \times 120 = 1200 $$

Similarly, the number of ways to choose exactly 3 boys and 2 girls is:

$$ \binom{5}{3} \times \binom{10}{2} = \frac{5!}{3!(5-3)!} \times \frac{10!}{2!(10-2)!} = 10 \times 45 = 450 $$

The number of ways to choose exactly 4 boys and 1 girl is:

$$ \binom{5}{4} \times \binom{10}{1} = \frac{5!}{4!(5-4)!} \times \frac{10!}{1!(10-1)!} = 5 \times 10 = 50 $$

And finally, the number of ways to choose exactly 5 boys is:

$$ \binom{5}{5} \times \binom{10}{0} = \frac{5!}{5!(5-5)!} \times \frac{10!}{0!(10-0)!} = 1 \times 1 = 1 $$

Therefore, the total number of favorable outcomes is:

$$ 1200 + 450 + 50 + 1 = 1701 $$

So, the probability that there are at least 2 boys when 5 students are chosen is:

$$ \frac{1701}{3003} \approx 0.5664 $$