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## Difficult Math Problem #95 - Probability

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800guy Master | Next Rank: 500 Posts
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#### Difficult Math Problem #95 - Probability

Fri Feb 09, 2007 10:32 am
from diff math problems doc. oa coming when some people answer with explanations..

From a group of 3 boys and 3 girls, 4 children are to be randomly selected. What is the probability that equal numbers of boys and girls will be selected?

A. 1/10

B. 4/9

C. 1/2

D. 3/5

E. 2/3

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### Top Member

Roland2rule Master | Next Rank: 500 Posts
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Sun Nov 05, 2017 11:09 am
probability of event n happening is given as
$$\left(\Pr\left(n\right)=\ \frac{\left(no.\ of\ required\ outcomes\right)}{\left(no.\ of\ possible\ outcomes\right)}\right)$$
In a group of 3 girls and 3 boys, equal number of boys and girls can be selected if 2 boys AND 2 girls are selected.
let x and y be the probability of selecting 2 boys and 3 boys and the probability of selecting 2 girls from 3 girls, respectively.
$$\Pr\left(x\right)=\ \frac{\left(no.\ of\ boys\ required\right)}{\left(total\ no.\ of\ boys\right)}=\frac{2}{3}$$
The same way;
$$\Pr\left(y\right)=\ \frac{\left(no.\ of\ girls\ required\right)}{\left(total\ no.\ of\ girls\right)}=\frac{2}{3}$$
therefore,
$$\Pr\left(x\ AND\ y\right)=\frac{2}{3}\cdot\frac{2}{3}=\ \frac{4}{9}$$

800guy Master | Next Rank: 500 Posts
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Mon Feb 12, 2007 3:25 pm
OA:

Total number of ways of selecting 4 children = 6C4 = 15

with equal boys and girls. => 2 boys and 2 girls. => 3C2 * 3C2 = 9.

Hence p = 9/15 = 3/5

### Top Member

Roland2rule Master | Next Rank: 500 Posts
Joined
30 Aug 2017
Posted:
405 messages
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Sun Nov 05, 2017 11:09 am
probability of event n happening is given as
$$\left(\Pr\left(n\right)=\ \frac{\left(no.\ of\ required\ outcomes\right)}{\left(no.\ of\ possible\ outcomes\right)}\right)$$
In a group of 3 girls and 3 boys, equal number of boys and girls can be selected if 2 boys AND 2 girls are selected.
let x and y be the probability of selecting 2 boys and 3 boys and the probability of selecting 2 girls from 3 girls, respectively.
$$\Pr\left(x\right)=\ \frac{\left(no.\ of\ boys\ required\right)}{\left(total\ no.\ of\ boys\right)}=\frac{2}{3}$$
The same way;
$$\Pr\left(y\right)=\ \frac{\left(no.\ of\ girls\ required\right)}{\left(total\ no.\ of\ girls\right)}=\frac{2}{3}$$
therefore,
$$\Pr\left(x\ AND\ y\right)=\frac{2}{3}\cdot\frac{2}{3}=\ \frac{4}{9}$$

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