SOLUTION: In a jar, you have 13 marbles: 6 blue, 4 green and 3 red. What is the problababilty that if you select 3 marbles, they're all green? Use permutation method, not conditional probabi

Question 661184: In a jar, you have 13 marbles: 6 blue, 4 green and 3 red. What is the problababilty that if you select 3 marbles, they're all green? Use permutation method, not conditional probabilty.
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
2 ways that i know of to solve this.
first way is the straight probability way.
no replacement is assumed.
on the first draw the probability of getting a green is 4/13.
on the second draw the probability of getting a green is 3/12.
on the third draw the probability of getting a green is 2/11.
the probability of getting a green on all 3 tries is 4/13 * 3/12 * 2/11 which is equal to 24 / 1716 which reduces to 2/143 which has a decimal equivalent of .0140.
the second way is using combination formulas.
the number of ways of getting 3 balls out of 13 is C(13,3) = 286.
the number of ways of getting 3 green balls out of 4 is C(4,3) = 4
the probability of getting 3 green balls out of 13 is equal to the number of ways you can get 3 green balls from 13 divided by the number of ways you can get 3 of any color ball from 13.
this becomes C(4,3) / C(13,3) which becomes 4/268 which becomes 2/143 which has a decimal equivalent of .0140.
both ways get the same answer.
both methods are valid.
C(n,x) is the combination formula to get x things out of n things.
the formula is actually C(n,x) = n! / ((n-x)! * x!)
permutations are not involved in this kind of problem.
permutations assume the sets are ordered in a certain way.
in this problem order was not important.
the number of combinations will be less than the number of permutations.
the formula for permutations is P(n,x) = n! / (n-x)!

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