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Difficult Math Problem #95 - Probability

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Difficult Math Problem #95 - Probability

by 800guy » 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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WHAT'S THE PROBABILITY TO HAVE GROUP ( 2 Boys & 2 girls)

by banona » Fri Feb 09, 2007 12:30 pm
I think answer is (3/5). Why ?

We have a group of 3 boys and 3 girls, 4 children are to be randomly selected.
all possible outcomes are of number of combinaison of sets of 4 members from a set of six numbers = 6!/2!*4! = 15

The assess the probability that an event of equal numbers of boys and girls will be selected, we have to count the number of possible sets that countain 2 boys and two girls;
there are many ways to do such computation, I would go for the following one using combinaisons and multiplication. Rephrase : We have a number of event sets equal to [( number of sets of two girls from 3)*( number of sets of two boys from 3)] = 3*3 = 9

Therfore probability is : 9/15 = 3/5

I hope there I was clear;
looking forward to reading some feed-backs

Banona

A. 1/10

B. 4/9

C. 1/2

D. 3/5

E. 2/3
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OA

by 800guy » 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
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by BTGmoderatorRO » 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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by Scott@TargetTestPrep » Tue Oct 29, 2019 6:02 am
800guy wrote: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

We are given that from a group of 3 boys and 3 girls, 4 children are to be randomly selected. We need to determine the probability that equal numbers of boys and girls will be selected, that is, the probability that two boys and two girls are selected.

We can use combinations to determine the number of favorable outcomes (that 2 boys and 2 girls are selected) and the total number of outcomes (that 4 children are selected from 6 children).

Let's first determine the number of ways we can select 2 boys from 3 boys and 2 girls from 3 girls.

# of ways to select 2 boys from a total of 3 boys: 3C2 = 3

# of ways to select 2 girls from a total of 3 girls: 3C2 = 3

Thus, the number of ways to select 2 girls and 2 boys = 3 x 3 = 9.

Now we can determine the total number of ways to select 4 children from a total of 6 children.

6C4 = (6 x 5 x 4 x 3)/(4 x 3 x 2 x 1) = 3 x 5 = 15

Thus, the probability of selecting an equal number of girls and boys is 9/15 = 3/5.

Answer: D

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