The Blair family was one of the first to come to the original 13 colonies (now part of the USA). They had 4 children. Assuming that the probability of a child being a girl is 0.5, find the probability that the Blair family had... (a) ...at least 3 girls

31.25% probability that the Blair family had at least 3 girls

Step-by-step explanation:

For each children, there are only two possible outcomes. Either it was a girl, or it was not. The probability of a child being a girl is independent of any other children. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

In which is the number of different combinations of x objects from a set of n elements, given by the following formula.

And p is the probability of X happening.

They had 4 children.

This means that

The probability of a child being a girl is 0.5

This means that

Probability of at least 3 children:

31.25% probability that the Blair family had at least 3 girls

hello from mrbilldoesmath!

answer:   (7x + 6y) ( 49x^2 - 42xy + 36y^2)

discussion:

in general, the factorization of a^3 + b^3 is

a^3 + b^ 3 = (a + b) ( a^2 - ab + b^2)

in our case note that 343 = 7^3 and 216 = 6^3 so the equation becomes:

343x^3 + 216y^3 = (7x)^3 + (6y)^3

if you think of a (above) as 7x and b as 6y the factorization becomes

343x^3 + 216y^3 =

((7x + 6y) (   (7x)^2 - 7x*6y + (6y)^2)   =

(7x + 6y) ( 49x^2 - 42xy + 36y^2)

you,

mrb

(a + b) (a^2 - a b + b^2)