# A committee of 6 teachers is to be formed from 5 male teachers and 8 female teachers. If the committee is selected at random, what is the probability that it has an equal number of male and female teachers?

A)$\dfrac{1}{{10}}$

B)$\dfrac{{140}}{{429}}$

C) $\dfrac{{150}}{{429}}$

D) $\dfrac{{160}}{{429}}$

E) $\dfrac{{170}}{{429}}$

Answer

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**Hint:**First we will know the meaning of the terms permutation and combination which are the important topics in probability.

In Probability, the term permutation refers to the process of arranging the outcomes in the order. Here, the order should be followed to arrange the items.

In Probability, the term combination refers to the process of selecting the outcomes in which it does not matter. Here, the order is not followed to arrange the items.

**Formula to be used:**

The formula to calculate the permutation is as follows.

\[{}^n{P_r} = n(n - 1)(n - 2).......(n - r + 1)\]

$ = \dfrac{{n!}}{{(n - r)!}}$ (! Is a mathematical symbol called the factorial)

Where, $n$ denotes the number of objects from which the permutation is formed and $r$ denotes the number of objects used to form the permutation.

The formula to calculate the combination is as follows.

${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$

Where, $n$ denotes the number of objects from which the combination is formed and $r$denotes the number of objects used to form the combination.

**Complete step-by-step solution:**

Our question is to select $6$ teachers from a set of $5$ male teachers and $8$ female teachers. Here we have to select the outcomes so that we are using the technique of combination.

And our exact question is to select $6$ teachers such that the males and females are equal.

First, we shall find the number of ways in which $6$ teachers can be selected from a set of $5$ male teachers and $8$ female teachers.

That is, \[{}^5{C_5}{}^8{C_1} + {}^5{C_4}{}^8{C_2} + {}^5{C_3}{}^8{C_3} + {}^5{C_2}{}^8{C_4} + {}^5{C_1}{}^8{C_{5}} + {}^5{C_0}{}^8{C_6} = 1716\]----$\left( 1 \right)$

Next we need to select $6$ teachers such that the males and females are equal.

That is, \[{}^5{C_3}{}^8{C_3} = 560\]----$\left( 2 \right)$

We know the formula of probability as,

$\text{Probability} =\dfrac{\text{Favourable Outcome}}{\text{Total Outcome}}$

So,

$\text{Probability} =\dfrac{560}{1716}$

$\text{Probability} =\dfrac{140}{429}$ (Cancelling numerator and denominator by 4)

**Hence, the required probability will be $\dfrac{{140}}{{429}}$ . (Dividing $\left( 2 \right)$ by $\left( 1 \right)$)**

**Note:**In Probability, it is the fraction of favourable outcome and total outcome. Here our total outcome is forming the committee by selecting 6 teachers out of 13(5-Male+8-Female) and the favourable outcome is 6 teachers(3-Female+3-Male). so we use the formula of combination as ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$

So our favourable outcome is ${}^5{C_3}\times {}^8{C_3}= \dfrac{{5!}}{{3!\left( {5-3}\right)!}} \times\dfrac{{8!}}{{3!\left( {8-3}\right)!}}$= $560$, and total outcome is ${}^{13}{C_6} = \dfrac{{{13}!}}{{6!\left( {13 - 6} \right)!}} = 1716$

Now the probability will be $= \dfrac{\text{Favourable Outcome}}{\text{Total Outcome}}$

$=\dfrac{560}{1716}$

$=\dfrac{140}{429}$ (Cancelling numerator and denominator by 4)

Hence option B is the correct answer.

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