2. A family has 5 natural children and two adopted girls. Suppose the probabilit
ID: 2933398 • Letter: 2
Question
2. A family has 5 natural children and two adopted girls. Suppose the probability of each natural child to be boy or girl is equal, and independent with each other. Let X be the number of girls of the seven children they have, (a) calculate the probability mass function of X (b) Now we generalize this question. Suppose a family has a natural children and b adopted girls. Suppose the probability of each natural child to be boy or girl is equal, and independent with each other. Let X be the number of girls of the a + b children they have, write out the probability mass function of X.Explanation / Answer
Solution
Let X = number of girls out of 7 (natural + adopted) children and Y = number of girls out of 5 natural children.
Part (a)
By the given conditions, Y ~ B(5, ½ ) and hence P(Y = y) = 5Cy(½)5 = (5Cy)/32 for all y = 0(1)5.
Given there are 2 adopted girls, minimum value of X is 2.
Pmf of X
X
Y
f(x)
0
-
0
1
-
0
2
0
(5C0)/32 = 1/32
3
1
(5C1)/32 = 5/32
4
2
(5C2)/32 = 10/32
5
3
(5C3)/32 = 10/32
6
4
(5C4)/32 = 5/32
7
5
(5C5)/32 = 1/32
Total
-
322/32 = 1
Or in general, f(x) = (5Cx - 2)/32, x = 0 (1) 7 [noting (nCx) = 0 for x < 0]
DONE
Part (b)
With a natural children and b adopted girls, the general form of pmf, f(x) would be
f(x) = (aCx - b)/2a, x = 0 (1) (a + b) [noting (nCx) = 0 for x < 0]
DONE
X
Y
f(x)
0
-
0
1
-
0
2
0
(5C0)/32 = 1/32
3
1
(5C1)/32 = 5/32
4
2
(5C2)/32 = 10/32
5
3
(5C3)/32 = 10/32
6
4
(5C4)/32 = 5/32
7
5
(5C5)/32 = 1/32
Total
-
322/32 = 1
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