A couple is starting a family and decides to have four children. Assume that the probability of having a boy or have a girl are equal. a) Find a probability model for the number of girls the couple has. b) Find the probability that the couple has at least one girl. c) What is the probability that the couple has exactly one boy or exactly one girl

0.9375 ; 0.5

Step-by-step explanation:

Given that :

p = 0.5

Probability model for the Number of girls the couple has :

P(x =x) = nCx * p^x * (1 - p)^(n - x)

P(x = 0) = 4C0 * 0.5^0 * 0.5^4 = 0.0625

P(x = 1) = 4C1 * 0.5^1 * 0.5^3 = 0.25

P(x = 2) = 4C2 * 0.5^2 * 0.5^2 = 0.375

P(x = 3) = 4C3 * 0.5^3 * 0.5^1 = 0.25

P(x = 4) = 4C4 * 0.5^4 * 0.5^0 = 0.0625

Probability of having atleast one girl

Number of trials, n = 4

P(x ≥ 1) = p(x = 1) + p(x = 2) + p(x = 3) + p(x = 4)

Using the relation :

P(x =x) = nCx * p^x * (1 - p)^(n - x)

Using calculator ;

P(x ≥ 1) = 0.9375

Probability of having exactly one boy or exactly one girl

1 boy 3 girls OR 1 girl 3 boys

P(x = 1)

P(x =x) = nCx * p^x * (1 - p)^(n - x)

P(x = 1) = 4C1 * 0.5^1 * 0.5^3

P(x = 1) = 4 * 0.5 * 0.125

P(x = 1) = 0.25

The probability of having one girl, 3 boys and vice versa is 0. 25

Hence, Probability of having exactly one boy or exactly one girl

0.25 + 0.25 = 0.5

for b its (t) and c its (d)

step-by-step explanation:

oof cheaters cheat because they’re lazy that’s why i cheat xd

: )

step-by-step explanation: