reto wrote:

Bunuel wrote:

reto wrote:

Private Benjamin is a member of a squad of 10 soldiers, which must volunteer 4 of its members for latrine duty. If the members of the latrine patrol are chosen randomly, what is the probability that private Benjamin will be chosen for latrine duty?

A. 1/10

B. 1/5

C. 2/5

D. 3/5

E. 4/5

Direct approach:\(\frac{C^1_1*C^3_9}{C^4_{10}}=\frac{2}{5}\).

Reverse approach:1 - (the probability of 4-member groups without Benjamin) = \(1 -\frac{C^4_9}{C^4_{10}}=\frac{2}{5}\).

Answer: C.

Is there a simple way to reduce \(\frac{C^1_1*C^3_9}{C^4_{10}}\) to 2/5 or do you also have to split it up to 9*8*7/3*2*1 / 210 and then go from there...?

you can look it at like this:

Remove Benjamin from the equation for now. You have 10 people from which only 9 are applicable to choose 4 volunteers from.

Thus the probabilities of selecting 1st,2nd , 3rd and 4th volunteers will be : (9/10) [you have 9 favorable choices out of 10 available], (8/9)[you have 8 favorable choices out of 9 available], (7/8)[you have 7 favorable choices out of 8 available], (6/7)[you have 6 favorable choices out of 17 available] .

The final probability (without Benjamin) will be (9/10)*(8/9)*(7/8)*(6/7) = 3/5.

Thus the Probability with Benjamin selected = 1-3/5 = 2/5.

As far as reducing nCr is concerned, there is only 1 formula: nCr = n!/ [r!*(n-r)!)]