Suppose that the probability of giving birth to a boy and the probability of giving birth to a girl are both 0.5. find the probability that in a family of seven children, the following occurs (a) all seven children are boys (b) there are two girls and five boys. (c) the youngest child is a girl (d) the oldest child is a girl. (a) the probability is (type an integer or a simplified fraction) (b) the probability is (type an integer or a simplified fraction.)

a)There is a 0.78% probability of all children being boys.

b) There is a 16.38% probability that there are two girls and five boys.

c) There is a 50% probability that the youngest child is a girl

d) There is a 50% probability that the oldest child is a girl

Step-by-step explanation:

Probability:

What you want to happen is the desired outcome.

Everything that can happen iis the total outcomes.

The probability is the division of the number of possible outcomes by the number of total outcomes.

In our problem, there is:

-50% probability of giving birth to a boy.

-50% probability of giving birth to a girl.

There are the following babies:

B1 - B2 - B3 - B4 - B5 - B6 - B7

(a) All seven children are boys

For each children there are a 50% probability of it being a boy. So:

There is a 0.78% probability of all children being boys.

(b) There are two girls and five boys.

There is a 50% probability that each children is a boy and 50% that each children is a girl.

We want five boys and two girls, so we have to permutate these probabilities. It is a permutation of 7 elements(all the children), with 5(boys) and 2(girls) repetitions. So:

There is a 16.38% that there are two girls and five boys.

(c) The youngest child is a girl

The probabilities are independent, so it is just the probability that B7 is a girl, that is 50%.

(d) The oldest child is a girl

Same as (c), just that the child is B1. The probability of B1 being a girl is 50%.

option c) graph of line going through 1, 4 and 3, 10

step-by-step explanation:

let y equal to margo's coin collection

let x equal to number of years

at the begining x=0, y=1 → point=(x,y)→point=(0,1)

she adds 3 coins per year, then y=1+3x

at x=1, y=1+3(1)=1+3→y=4→point=(x,y)→point=(1,4)

at x=3, y=1+3(3)=1+9→y=10→point=(x,y)→point=(3,10)

the graph of line going through (1, 4) and (3, 10), then the option c is the correct.