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12.11.2020 •
Mathematics
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?
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Ответ:
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)
Ответ:
Answer not good i caint see da out side world or my friends
Step-by-step explanation:
idk your answer sorry Btw how r u