Answer: 0.75
Step-by-step explanation:
Given : A total of 10 students are entered in a drawing to win 3 identical prizes.
The group of students is composed of 6 boys and 4 girls.
We know that the number of combinations of n things taking r at a time is given by :-
[tex]^nC_r=\dfrac{n!}{(n-r)!r!}[/tex]
The number of ways to choose any 3 out of 10 to win prizes will be :-
[tex]^{10}C_r=\dfrac{10!}{(10-3)!3!}=\dfrac{10\times9\times8\times7!}{7!\times6}=120[/tex] (1)
The number of ways to choose 2 boys and 1 girl to win prizes will be :-
[tex]^6C_2\times^4C_2=\dfrac{6!}{2!4!}\times\dfrac{4!}{2!2!}\\\\=15\times6=90[/tex] (2)
Now, the probability that 2 boys and 1 girl win prizes will be :-
[tex]\dfrac{90}{120}=0.75[/tex] [ Divide (2) by (1) ]
Hence, the required probability = 0.75
