Question

There are 3 directors, 4 management executives, 2 statisticians and 1 administrator. A committee of 4 members among them is to be formed. Find the probability that the committee consists of i) one of each kind III) one administrator and the other three III) at least one director
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Answer 1

Answer:

1) 4/35

2) 11/14

3) 5/6

Step-by-step explanation:

First find the number of selections possible.

$\frac{10!}{(10-4)!4!} = 210$

There are 210 possible selections. Now we can determine the possibility for each group. For the first probability, since we only want one member from each committee, you would create a fraction by (group members in total / group member being extracted). For this instance, since its only one, we will do whole numbers. Multiply each other.

$4*3*2*1=24$

The probability for each member to get selected from each kind is : $24/210$ or can be simplified to $4/35$.

The probability of (I assume) at least one administrator is slightly different. Separate our two selections, the other three and the administrator. Find the number of selections possible for the other three... 4+3+2 = 9 others.

Find the probability for both a administrator included and three others.

$\frac{11}{3} * \frac{1}{1} \\\frac{11!}{(11-3)!3!} \\\\165$

$\frac{165}{210} = \frac{11}{14}$

The probability for an administrator and the other three is 165/210 or 11/14.

For the last one, you will figure out what the probability is for no director and use that % to substract from 100.

4+2+1 = 7

Get selections possible.

$\frac{7!}{(7-4)!4!} = 35$

Get probability of no director.

$\frac{35}{210} = \frac{1}{6}$

Substract 1/6 from 100%

1 - 1/6 = 5/6

The probability of having at least one director is 5/6.