Question 1198349: Assume that, when a child is born, the probability it is a girl is ½ and that the sex of the child does not depend on the sex of an older sibling. Find the probability distribution for the number of girls in a family with 4 children
Found 2 solutions by ewatrrr, math_tutor2020:
Answer by ewatrrr(24785)   (Show Source):
You can put this solution on YOUR website! the probability it is a girl is ½
Find the probability distribution for the number of girls in a family with 4 children
(1/2)(1/2)(1/2)(1/2) = 1/16
Answer by math_tutor2020(3817)  (Show Source):
You can put this solution on YOUR website!
It appears that the tutor @ewatrrr might have misread the question.
The task is to find the probability distribution for the number of girls in this family. The teacher wants a probability table showing all the possible outcomes, and associated probabilities.
Convention is to use the binomial distribution. However, I'll take a different approach.
X = number of girls
X ranges from X = 0 to X = 4
The calculation @ewatrrr has provided applies to both X = 0 girls (aka 4 boys) and also X = 4 girls (0 boys) due to symmetry.
B = boy
G = girl
X = 0 girls
BBBB
X = 1 girl
BBBG
BBGB
BGBB
GBBB
There are four cases where the family has 1 girl
We can see this using the nCr formula with n = 4 and r = 1.
X = 2 girls
BBGG
BGGB
BGBG
GBBG
GBGB
GGBB
There are six cases where the family has 2 girls
This value (6) can be calculated by computing the nCr value when n = 4 and r = 2.
X = 3 girls
GGGB
GGBG
GBGG
BGGG
There are four cases where the family has 3 girls (notice the similar structure to the X = 1 girl case)
This value (4) can be calculated by computing the nCr value when n = 4 and r = 3.
X = 4 girls
GGGG
There is one case where the family has 3 girls (notice the similar structure to the X = 0 girls case)
There's no need to use nCr here, but if you wanted then it would be n = 4 and r = 4.
There are 2^4 = 16 ways to have four kids.
This is the probability distribution
| X | P(X) | | 0 | 1/16 | | 1 | 4/16 | | 2 | 6/16 | | 3 | 4/16 | | 4 | 1/16 |
Thing to notice: The numerators 1,4,6,4,1 are found in the same row of Pascal's Triangle
I decided to not reduce the fractions to keep the denominators consistent.
This is what it looks like after reducing those fractions
| X | P(X) | | 0 | 1/16 | | 1 | 1/4 | | 2 | 3/8 | | 3 | 1/4 | | 4 | 1/16 |
Another thing to notice: The P(X) values add to 1.
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