A group of individuals containingbboys andggirls is lined up in a randomorder; that is, each of the (g+b)! permutations is assumed to be equally likely. What is the probability that the person in theithposition (1≤i≤b+g) isa girl?

g/(b + g)

Step-by-step explanation:

We are told the group of individuals contain b number of boys and g number of girls.

Now the number of ways of selecting 1 girl out of g number of girls is;

gC1 = g!/(g - 1)! = (g × (g - 1)!)/(g - 1)!

(g - 1)! will cancel out to give;

gC1 = g

Now, since total number of boys and girls is b + g, and we have chosen one girl out, the remaining number of people will be; b + g - 1.

Thus, number of ways to arrange the remaining people will be;

(b + g - 1)!

Now, the probability that the person in the ith position (1 ≤ I ≤ b+g) is a girl will be;

(gC1 × (b + g - 1)!)/(b + g)!

We know that gC1 = g from earlier derivation.

Expanding this gives;

[g(b + g - 1) × (b + g - 2)!]/[(b + g) × (b + g - 1) × (b + g - 2)!]

(b + g - 1) × (b + g - 2)! will cancel out to give;

g/(b + g)

28

step-by-step explanation:

8+8+6+6

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