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A family decides to have children until it has three children of the same gender. Assuming...

A family decides to have children until it has three children of the same gender. Assuming P(B) = P(G) = 0.5, what is the pmf of X = the number of children in the family? x 0 1 2 3 4 5 6
2 0
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Answer #1
Concepts and reason

A discrete probability distribution lists each possible value the random variable can assume, together with its probability.

If the random variable X is a one-dimensional discrete random variable taking at most a countable infinite number of values then its probabilistic behaviour at each real point is described by a function called the probability mass function.

It is denoted as,

Р(х 3Dх)

From the given information, let the random variable X indicates the number of components meet specifications. Since, there are two components in the system X can take the values 0 or 1 or 2.

Fundamentals

A discrete probability distribution must satisfy the following conditions.

1.The probability of each value of the discrete random variable is between 0 and 1, inclusive.

0s P(X x)s1

2.The sum of all the probabilities is 1.

ΣΡ(Χ=3)=1

The probability of having a boy or girl is, Р(B) -
P(G) 0.5
.

The family decides to have children until it has three children of the same gender.

The probability of having 0 or 1 or 2 children is,

P(0) P() P(2) 0

The possibilities of having 3 children are,

GGG, GGB,GBG,BGG, GBB, BGB, BBG,BBB

The probability of having 3 children of same gender is,

2
P(3 children)
1
4

The possibilities of having 4 children are,

GGGG,GGGB,GGBG,GGBB
GBGG,GBGB,GBBG,GBBB,
BGGG,BGGB,BGBG,BGBB,
BBGG,BBGB,BBBG,BBBB

The outcomes that satisfy the condition are,

GGBG,GBGG, BGGG,
BBGB, BGBB,GBBB

The probability of having 4 children in which 3 are of same gender is,

6
P(4) 16
3

The probability of having 5 children in which 3 are of same gender is,

P(5) 1-P(3)-P(4)
1 3
=1-
4 8
=1-
8

Ans:

The frequency distribution is,

0 1234 56
5 6
х
33
0
1
p(x) 000
88
4

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Answer #2

The only possible values of X are 3,4, and 5.

P(3) = P(X = 3) = P(first 3 are B’s or first 3 are G’s) = 2(.5)3 = .250

above is bionomial with n = 3 and p = 1/2

P(x=3 boys) = 3 C 3 0.5^3 0.5^0 = 1/8
P(x=3 girls) = 3 C 3 0.5^3 0.5^0 = 1/8

adding = 1/8+1/8 = 2/8 =1/4 =0.25

P(4) = P(two among the 1st three are B’s and the 4th is a B) + P(two among the 1st three are G’s and the 4th is a G) =

2* 3 C 2 0.5^4 = 0.375


P(5) = 1 – p(3) – p(4) = .375

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