A couple plans to have children until they get a girl, but they agree they will not have more than three children, even if all are boys. Assume that the probability of having a girl is 47.00%. a) Create a probability model for the number of children they'll have. b) Find the expected number of children. c) Find the expected number of boys they'll have. a) X P(Xequalsx) 1 nothing 2 nothing 3 nothing (Round to four decimal places as needed.) b) E(X)equals nothing (Round to four decimal places as needed.) c) E(number of boys)equals nothing

a)Let X be number of children

X=1,2,3

P(X=1)= 0.47

P(X=2) = 0.2491

P(X=3) = 0.942923

b) 3.7970

c) 1.6232

Step-by-step explanation:

The complete question is:

A couple plans to have children until they get a girl, but they agree they will not have more than three children, even if all are boys. Assume that the probability of having a girl is 47.00%.

a) Create a probability model for the number of children they'll have.

X=1,2,3

P(X=1)=??

P(X=2)= ??

P(X=30=???

(Round to four decimal places as needed

b) Find the expected number of children.

E(X)= ???

c) Find the expected number of boys they'll have.

Expected number of boys= ???

Solution:

Probability of a girl= 0.47

Probability of a boy= 0.53

a) P(X=1)= 0.47

P(X=2) = 0.47× 0.53= 0.2491

P(X=3)= 0.47× 0.53× 0.53 + 0.53× 0.53× 0.53

             = 0.942923

b) E(number of children)= 1× P(X=1) + 2 ×P(X=2) + 3 × (PX=3)

                                       = 3.796969

c) Y: number of boys

P(Y=1)= 0.53×0.47= 0.2491

P(Y=2) = 0.53×0.53×0.47=0.46375

P(Y=3)= 0.53× 0.53× 0.53= 0.148875

E(Y)= P(Y=1)×1 + P(Y=2)×2 + P(Y=3)×3

       = 0.148875×3 +0.46375×2+0.2491 ×1

      = 1.6232

ou would have 12 of each

there is no graph so i cant sorry

step-by-step explanation: