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A team of 6 members is formed from 6 males and 8 females.

What is the probability that the team has equal numbers of male and female ?

I calculated 20/3003 = 0.0067 Is it correct? Help me with this problem, please 🌱

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  • $\begingroup$ Where did $20$ and $3003$ come from? $\endgroup$ – Arthur Nov 13 '18 at 4:48
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    $\begingroup$ You forgot that the male committee members can be chosen in more than one way. $\endgroup$ – Fabio Somenzi Nov 13 '18 at 4:52
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    $\begingroup$ No, it's not like that. You had 3003 ways to pick 6 members out of 14, and 20 ways to pick 3 women out of 6. Now you need to account for the different ways to pick 3 men out of 8. Each choice of 3 men and 3 women gives you a valid committee. $\endgroup$ – Fabio Somenzi Nov 13 '18 at 5:10
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    $\begingroup$ No, each committee has both women and men. You use the so-called "multiplication rule." You'd use the "addition rule" if you were counting the different ways to pick either three women or three men. But here you need to pick 6 members, not 3. For every way to choose 3 women, there are 56 ways to choose 3 men and vice versa, for every way to choose 3 men, there are 20 ways to choose 3 women. $\endgroup$ – Fabio Somenzi Nov 13 '18 at 5:37
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    $\begingroup$ How many ways are there to form a committee of 6 choosing from 14? You correctly counted 3003. Now you need to count how many of those committees have 3 men and 3 women, and then take the ratio. $\endgroup$ – Fabio Somenzi Nov 13 '18 at 5:40
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You start with the equal number of male and female part in a team of 6 which gives you 3 male members and 3 female members. Now your job is to select 3 male team members from a group of 6 and 3 female members from a group of 8.

Number of ways to select male members = ${6}\choose 3$

Number of ways to select female members = ${8}\choose 3$

As they are independent events, we get the number of ways to form a team of equal male and female members by multiplying to get

$${{6}\choose{3}} {{8}\choose{3}}$$

And as we know total number of ways to select a team out of $8 + 6 = 14$ people is = $14\choose6$

Dividing the two, we get the probability as

$$\frac{{{6}\choose{3}} {{8}\choose{3}}}{14\choose6} = \frac{1120}{3003} = 0.3729$$

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There are $6M$ and $8F$.
For equal number of $M$ and $F$ in a team, you need $3M$ and $3F$.
So, you require to select $3M$ from $6$ and $3F$ from $8$.

So, selection goes like: $$^6C_3\cdot ^8C_3$$
Whereas, total ways to select $6$ people from $14$ is $$^{14}C_6$$
So, required probability will be:$$\frac{^6C_3 \cdot ^8C_3}{^{14}C_6}$$

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