A foreman in a manufacturing plant has three men and three women working for him. He needs to choose two workers for a special job. Not wishing to show any biases, he decides to select two workers at random. If the random variable Y denotes the number of females chosen, find the probability distribution of Y.

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There are 3 male workers and 3 female workers and 2 are to be selected without bias, ie at random.

The number of ways of choosing 2 workers out of 6 is

`""^6C_2 = (6!)/(2!4!) = (6.5)/(2.1) = 15`

If two females are chosen, there are 3 ways of choosing a particular pair of females out of the three available females. Similarly if two males are chosen, there are 3 ways of choosing a particular pair of males out of the three available males.

Therefore ` `

**P[Y=0] = P[Y=2] = 3/15 = 1/5 and P[Y=1] = 1 - 3/15 - 3/15 = 9/15 = 3/5 **

There are 3 male workers and 3 female workers and 2 are to be selected without bias, ie at random.

Since there are equal numbers of males and females then the probability of choosing a male or a female is the same. Therefore the probability of any combination of males and females is equal.

The number of ways of choosing 2 workers out of 6 is

`""^6C_2 = (6!)/(2!4!) = (6.5)/(2.1) = 15`

**The probability of any of Y=0, Y=1, Y=2 is then 1/15**

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