SOLUTION: Suppose S={a,b,c,d,e} . If a three-element subset of S is selected at random, what is the probability it will contain one but not both of the elements b and c? Enter your an

Question 1122708: Suppose S={a,b,c,d,e} .
If a three-element subset of S is selected at random, what is the probability it will contain one but not both of the elements b and c?

Enter your answer as a whole number or a fraction in lowest terms.
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
it can contain b or c, but not both.

i get 6/10 using brute force and then confirmed by formula.

brute force method.

a,b,c,d,e
sets of 3 are:

a,b,c
a,b,d
a,b,e
a,c,d
a,c,e
a,d,e
b,c,d
b,c,e
b,d,e
c,d,e

that's 10 possible choices of getting 3 elements out of 5 where order is not important.

out of those, remove the ones that have both c and c in them.

those are:

a,b,c *****
a,b,d
a,b,e
a,c,d
a,c,e
a,d,e
b,c,d *****
b,c,e *****
b,d,e
c,d,e

you will be left with:

a,b,d
a,b,e
a,c,d
a,c,e
a,d,e
b,d,e
c,d,e

now remove the ones that don't have b or don't have c in them.

the are:

a,b,d
a,b,e
a,c,d
a,c,e
a,d,e *****
b,d,e
c,d,e

you will be left with:

a,b,d
a,b,e
a,c,d
a,c,e
b,d,e
c,d,e

that's 6 out of the original 10 possibilities.

therefore, the probability you will get b or c, but not both, is 6 out of 10.

by formula, you can do it a couple of ways.

the first way is:

probability of getting b only would be 1/5 * 3/4 * 2/3 * c(3,1) = 6/60 * 3 = 18/60

probability of getting c only would be 1/5 * 3/4 * 2/3 * c)3,1) = 6/60 * 3 = 18/60

the probability of getting b or c only would therefore be 18/60 + 18/60 = 36/60 = 6/10.

the formula works as follows;

the probability of getting b only on the first draw is 1/5.
the probability of not getting c on the second draw is 3/4.
the probability of not getting c on the third draw is 2/3.
that gets you a probaiblity of 6/60 = 1/10.
this can happen in 3 ways, i.e. b on the first draw, b on the second draw, b on the third draw, therefore 1/10 * 3 = 3/10.

same thing happens with c.

the other way is to use combination theory to get:

the number of ways you can get b only out of 3 is c(3,1).
the number of ways you can get 3 out of 5 is c(5,3).

the probability is therefore c(3,1) / c(5,3) = 3/10

same probability to get c only.

probability of b or c is therefore 3/10 + 3/10 = 6/10

the c(3,1) means 3 ways to get b only.
you would get b,x,x, or x,b,x or x,x,b
same goes for c.

c(n,x) = n! / (x! * (n-x)!)

c(3,1) is therefore 3! / (1!*2!) = (3*2*1) / (1*2*1) = 3

c(5,3) is therefore 5! / (3! * 2!) = (5*4*3!) / (3!*2*1) = (5*4)/2 = 10

all 3 ways gets you the same result.

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