Question

The Jones family was one of the first to come to the U.S. They had 4 children. Assuming that the probability of a child being a girl is .5, find the probability that the Jones family had:_______________.
(a) at least 3 girls
(b) at most 3 girls

Answer 1

Answer:

a= 1/4

b=15/16

Explanation:

There are no sequences or anything that can differ the 4 children, so the order in this probability is not important. Let's say chance the chance for a girl is A and chance for a boy is B, then  

P(A)= 0.5 = 50%

P(B)= 1-0.5= 50%

(a) at least 3 girls

At least 3 girls mean you need to have 3 girls or more. The events that can fulfill at least 3 girls out of 4 children are  

1. 3 girls and 1 boy

2. 4 girls and 0 boy

We can find the probability by adding these 2 events. The calculation will be:

P(3A1B) + P(4A0B) = 3C1 * A^3 * B^1          +           4C0 * A^4  * B^0

P(3A1B) + P(4A0B) = 3 * 0.5^3  * 0.5^1       +         1 * 0.5^4 * 0.5^0

P(3A1B) + P(4A0B) = 3/16 + 1/16= 4/16 = 1/4

(b) at most 3 girls

At most 3 girls mean that less or equal to 3 girls. There are only 4 children, so the only event that did not fulfill the condition is 4 girls and 0 boy. It is easier to find the probability for 4 girls and 0 boy then calculate the negation, it will be:

P(4A0B)= 4C0 * A^4  * B^0=

P(4A0B)=  1 * 0.5^4 * 0.5^0=

P(4A0B)= 1/16

The probability will be ~P(4A0B)= 1- 1/16= 15/16