Question 1132436: Here’s a question we were unsure of.
According to a study of the numbers of male and female children in families in a certain population, a model has been determined that for any subsequent children, the probability that they will be of the same gender as the previous child is 3/5. Using this model, calculate the probability that, in a randomly chosen family of 3 children, there will be;
a) 2 males and 1 female
b) at least one female
Hence, construct the probability distribution table of X if X is the number of female selected among the 3 children.
Found 2 solutions by Boreal, greenestamps:
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website!
the eight outcomes are, and to answer the question properly, X goes in place of "F:
MMM--0.18 from 0.5*0.6*0.6
MMF--0.12 from 0.5*0.6*0.4
MFM--0.08 from 0.5*0.4*0.4
MFF--0.12 from 0.5*0.4*0.6
FFF--0.18
FFM--0.12
FMM--0.12
FMF--0.08
They add to 1.00
for 2M 1F the probability is 0.32
for at least one F is 0.82
Answer by greenestamps(13209)  (Show Source):
You can put this solution on YOUR website!
We need to assume that there is equal probability that the first child is male or female. After that, the probability is 3/5 that the next child will be the same gender and 2/5 that it will be different. Then...
(1) all male
P(MMM) = (1/2)(3/5)(3/5) = 9/50
(2) 2 male, 1 female
P(MMF) = (1/2)(3/5)(2/5) = 6/50
P(MFM) = (1/2)(2/5)(2/5) = 4/50
P(FMM) = (1/2)(2/5)(3/5) = 6/50
total 16/50
(3) 1 male, 2 female
P(MFF) = (1/2)(2/5)(3/5) = 6/50
P(FMF) = (1/2)(2/5)(2/5) = 4/50
P(FFM) = (1/2)(3/5)(2/5) = 6/50
total 16/50
(4) all female
P(FFF) = (1/2)(3/5)(3/5) = 9/50
Note that the sum of all the probabilities is 1, giving confidence that the calculations were correct.
You can use the probabilities shown to answer the questions that were asked.
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