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Mars2501 [29]
3 years ago
8

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.
Mathematics
2 answers:
Papessa [141]3 years ago
4 0

Answer:

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

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

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

Step-by-step 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)}=\frac{6}{8}

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

3) P(at least two girls)

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

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

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

Anni [7]3 years ago
4 0
 

<span>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</span>

<span>2) BBG

3) BGB</span>

<span>4) BGG

5) GBB</span>

6) GBG

<span>
7) GGB</span>

<span>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</span>

P = 4/8 = 1/2

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