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Let X denote the random variable given by the number of boys in a family of four children. Find the probability distribution if a girl is as likely as a boy. (Round your answers to four decimal places.)

Random variable, x 0 1 2 3 4 5 6
P(X=x)

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The probability distribution is:

P(X=0)=0.0625

P(X=1)=0.5

P(X=2)=0.375

P(X=3)=0.25

P(X=4)=0.0625

In a family of four children where a girl is as likely as a boy, the possible values for the number of boys (random variable X) can range from 0 to 4. We can use the binomial probability distribution to find the probabilities.

The probability mass function (PMF) for a binomial distribution is given by:

P(X=k)=( n k )p^k (1−p)^n−k

where:

n is the number of trials (number of children in this case),

k is the number of successes (number of boys in this case

p is the probability of success on each trial (probability of having a boy),

(n k)​ is the binomial coefficient, which represents the number of ways to choose

k successes from

n trials.

In this case,

n=4 (four children), and

p=0.5 (equal probability of having a boy or a girl).

Now, let's calculate the probabilities for each value of X:

P(X=0)=( 0 4 )(0.5)^0 (0.5)^4

P(X=1)=( 1 4 )(0.5)^1 (0.5)^3

P(X=2)=( 2 4 )(0.5)^2 (0.5)^2

P(X=3)=( 3 4 )(0.5)^3 (0.5)^1

P(X=4)=( 4 4 )(0.5)^4 (0.5)^0

Let's calculate these probabilities:

P(X=0)=( 0 4 )⋅1⋅0.0625=1⋅0.0625=0.0625

P(X=1)=( 1 4 )⋅0.5⋅0.125=4⋅0.125=0.5

P(X=2)=( 2 4 )⋅0.25⋅0.25=6⋅0.0625=0.375

P(X=3)=( 3 4 )⋅0.125⋅0.5=4⋅0.0625=0.25

P(X=4)=( 4 4 )⋅0.0625⋅1=1⋅0.0625=0.0625

So, the probability distribution is:

P(X=0)=0.0625

P(X=1)=0.5

P(X=2)=0.375

P(X=3)=0.25

P(X=4)=0.0625

User Dnur
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