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Probability of having atleast a girl= 1- Probability of having all boys

Probability of having a boy= 1/2
Probability of having three boys= (1/2)(1/2)(1/2) = 1/8

Probability of having atleast a girl = 1- (1/8) = 7/8 = 0.875

Correct Answer= B
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Bunuel
If a family has three children, what is the probability that the family has at least one girl?


A. 1

B. 0.875

C. 0.75

D. 0.50

E. 0.375

The two events “at least one girl” and “no girls” are complementary events, which means that the sum of their probabilities is 1. Thus, we have:

P(at least 1 girl) = 1 - P(no girls)

P(no girls) = 1/2 x 1/2 x 1/2 = ⅛

So, P(at least 1 girl) = 1 - P(no girls) = 1 - 1/8 = 7/8 = 0.875.

Answer: B
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Bunuel
If a family has three children, what is the probability that the family has at least one girl?


A. 1

B. 0.875

C. 0.75

D. 0.50

E. 0.375

Total sample space = BBB,BBG,BGB,GBB,GGB,GBG,BGG,GGG where G=Girl and B=BOY

probability that at least one of the children will be girl = 7/8 = .875
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Bunuel
If a family has three children, what is the probability that the family has at least one girl?


A. 1

B. 0.875

C. 0.75

D. 0.50

E. 0.375

We can use the formula:

P(at least one girl) = 1 - P(no girls)

P(no girls) = (1/2)^3 = 1/8.

P(at least one girl) = 1 - 1/8 = 7/8 = 0.875.

Answer: B
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