A married 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

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Answer:

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 = [tex](\frac{1}{2})^{2}[/tex] + [tex](\frac{1}{2})^{3}[/tex] + [tex](\frac{1}{2})^{3}[/tex]

i.e. Probability = [tex]\frac{1}{4} +\frac{1}{8} +\frac{1}{8}[/tex]

i.e. Probability = [tex]\frac{1}{4} +\frac{2}{8}[/tex]

i.e. Probability = [tex]\frac{1}{4} +\frac{1}{4}[/tex]

i.e. Probability = [tex]\frac{2}{4}[/tex]

i.e. Probability = [tex]\frac{1}{2}= 0.5[/tex]

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.

The probability that the couple has at least two girls is 1/2.

Given

A married 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.

What is probability?

The quality or state of being probable; the extent to which something is likely to happen or be the case:

Consider the probability of the couple having less than 2 girls.

The cases are as follows;

BGB, GBB, BB

Where B is a boy and G is a girl.

Then,

The probability is;

[tex]\rm P' = \left( \dfrac{1}{2} \right )^3+ \left( \dfrac{1}{2} \right )^3+ \left( \dfrac{1}{2} \right )^2\\\\P'= \dfrac{1}{8} +\dfrac{1}{8} +\dfrac{1}{4}\\\\P'= \dfrac{1+1++2}{8}\\\\P'=\dfrac{4}{8}\\\\P'=\dfrac{1}{2}[/tex]

Therefore

The probability that the couple has at least two girls is;

[tex]\rm P = 1-P'\\\\P = 1-\dfrac{1}{2}\\\\P = \dfrac{1}{2}[/tex]

Hence, the probability that the couple has at least two girls is 1/2.

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