Amarried couple agreed to continue bearing a new child until they get two boys , but not more than 4 children. assuming that each time that a child is born, the probability that is a boy is 0.5 independent from all other times. find the probability that the couple has atleast two girls. a. 1/2 b. 5/16 c. 5/8 d. 4/15

A. 1/2 = 0.5

Step-by-step explanation:

We are given that the couple stops bearing a child until they get two boys.

Now, if the couple bears less than 2 girls.

Then the possible cases are: BB, BGB and GBB.

As the probability of the child being a boy = 0.5

Thus, the probability of the child being a girl = 1 - 0.5 = 0.5.

Hence, the probability that the couple bears less than 2 girls = sum of probabilities of BB, BGB and GBB.

i.e. Probability of less than 2 girls = + +

i.e. Probability =

i.e. Probability =

i.e. Probability =

i.e. Probability =

i.e. Probability =

Thus, the probability of less than 2 girls = 0.5

So, the probability of atleast 2 girls = 1 - Probability of less than 2 girls

i.e. The probability of atleast 2 girls = 1 - 0.5 = 0.5

Hence, the probability that the couple has atleast 2 girls is 0.5 = 1/2.

i think it's either c or d

step-by-step explanation: the entire question is extremely odd. 04 obviously refers to the interest rate. honestly i will say the answer is c. because the (2 x 12) part would mean it's compounded semiannually. i'll say choose c.