Question

Mr. and Mrs. Romero are expecting triplets. Suppose the chance of each child being a boy is 50% and of being a girl is 50%. Find the probability of each event.

P(at least one boy anf one girl)

P(two boys and one girl)

P(at least two girls)

I don't care if you cab only answer one, I would very much appreciate it if you still answered that one instead of nothing.

Answer 1

Let's represent a boy with B and a girl with G

For example BBB is three boys and BGB is two boys and one girl.

These are all the possibilities

1) BBB

2) BBG

3) BGB

4) BGG

5) GBB

6) GBG

7) GGB

8) GGG

There is in total 8 different possibilities.

Now let's analyze the conditions

p (at least one boy and one girl)

That excludes only BBB and GGG so they are 6 possibilities out of 8

p = 6/8 = 3/4

p (two boys and one girl).

Those are 3 out of 8

p = 3/8

p (at least two girls)

Those are 3 (two girls) + 1(three girls) = 4

P = 4/8 = 1/2

Please, let me know if this was satisfactory for you.

Answer 2

1)
`\text{P(at least one boy and one girl)}=(3)/(4)`

2)
`\text{P(at least one boy and one girl)}=(3)/(8)`

3)
`\text{P(at least two girls)}=(1)/(2)`

Explanation:

Given : Mr. and Mrs. Romero are expecting triplets. Suppose the chance of each child being a boy is 50% and of being a girl is 50%.

To Find : The probability of each event.

1) P(at least one boy and one girl)

2) P(two boys and one girl)

3) P(at least two girls)

Solution :

Let's represent a boy with B and a girl with G

Mr. and Mrs. Romero are expecting triplets.

The possibility of having triplet is

BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG

Total outcome = 8

`\text{Probability}=\frac{\text{Favorable outcome}}{\text{Total number of outcome}}`

1) P(at least one boy and one girl)

Favorable outcome = BBG, BGB, BGG, GBB, GBG, GGB=6

`\text{P(at least one boy and one girl)}=(6)/(8)`

`\text{P(at least one boy and one girl)}=(3)/(4)`

2) P(at least one boy and one girl)

Favorable outcome = BBG, BGB, GBB=3

`\text{P(at least one boy and one girl)}=(3)/(8)`

3) P(at least two girls)

Favorable outcome = BGG, GBG, GGB, GGG=4

`\text{P(at least two girls)}=(4)/(8)`

`\text{P(at least two girls)}=(1)/(2)`