A couple plans to have no more than three children, and they will keep having children until they have a girl. So, if their first child is a girl, they will stop and have only one child. However, if their first child is a boy, they will try again and have a second child. As it turns out, the probability of having a boy is slightly greater than having a girl. Here is the probability distribution for the number of boys the couple could have. Boys 0 boys 1 boy 2 boys 3 boys Probability 0.490 0.250 0.127 0.133

What is the expected number of boys the couple will have? (Recall: the expected value is the mean). 0.903 1.5 0.228

0.903.

Step-by-step explanation:

Okay, from the question we are given the following information or data or parameters:

''couple plans to have no more than three children, and they will keep having children until they have a girl. So, if their first child is a girl, they will stop and have only one child"

Additionally, from the question we are given that; ''However, if their first child is a boy, they will try again and have a second child".

Also, the probability distribution is given as Boys 0, boys 1 , boy 2, boys 3 boys with Probability 0.490, 0.250, 0.127, 0.133 respectively.

That is, we now have:

Number of boys × probability.

Boys 0 = 0 × 0.490 = 0.

Boys 1 = 1 × 0.250 = 0.250.

Boys 2 = 2 × 0.127 = 0.254.

Boys 3 = 3 × 0.133 = 0.399.

The addition of the products for each Number of boys × probability gives us the expected value. That is to say;

0 + 0.250 + 0.254 + 0.399 = 0.903 = mean.

third option: 6x+8

step-by-step explanation:

we can see that the given question is a sum of series and the values of n vary from 0 to 3

the first term will be:

when n=0, [(0)*x+2] => (0+2) => 2

when n=1, [(1)*x+2] => (x+2) => x+2

when n=2, [(2)*x+2] => 2x+2

when n=4, [(3)*x+2] => 3x+2

so adding up all the terms will give us:

=2+x+2+2x+2+3x+2

=x+2x+3x+2+2+2+2

=6x+8

so,

third option is the correct answer ..