Respuesta :
Answer:
(c), 0.75
Step-by-step explanation:
It is given that, the probability of boy is same as the probability of girl.
Number of children family plans to have = 3
Consider, B be the event that represents that child is a boy
G be the event that represents that child is a girl.
The simple events for the provided case could be written as:
S = {(GGG, BBB, GBB, BGB, BBG, BBG, GBG, GGB)}
From above simple event, it is clear that
P(3 boys) = 1/8 and P(3 Girls) = 1/8
Thus, the probability of having at least one boy and at least one girl can be calculated as:
[tex]Probability = 1- (\frac{1}{8} + \frac{1}{8} ) = \frac{6}{8} =\frac{3}{4}[/tex]
Thus, the required probability is 0.75.
Hence, the correct option is (c), 0.75
Answer:
C. 0.75
Step-by-step explanation:
Total Possible 8 Outcomes : GGG, BGG, GBG, GGB ,BBB, GBB, BGB, BBG
Probability of having atleast one boy & atleast one girl = 1- (Probability of having all girls, all boys)
Pr (All girls, All boys) = GGG, BBB = 2 outcomes
So, Pr [Atleast one boy & ateast one girl] = 1 - 2/8
= 6/8
= 0.75