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,
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.
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.
2.The sum of all the probabilities is 1.
The probability of having a boy or girl is, .
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,
The possibilities of having 3 children are,
The probability of having 3 children of same gender is,
The possibilities of having 4 children are,
The outcomes that satisfy the condition are,
The probability of having 4 children in which 3 are of same gender is,
The probability of having 5 children in which 3 are of same gender is,
Ans:
The frequency distribution is,
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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