SOLUTION: I cannot seem to figure out how this problem is worked. Any help is appreciated. I Know the answer is 1/56 but I don't have a clue at the moment how that can be.
Here is the probl
Algebra.Com
Question 325063: I cannot seem to figure out how this problem is worked. Any help is appreciated. I Know the answer is 1/56 but I don't have a clue at the moment how that can be.
Here is the problem:
In a certain lottery, 3 different numbers between 1 and 8 inclusive are drawn at random. These are the winning number. If you choose 3 different numbers at random between 1 and 8, what is the probability you will match the winning numbers? Assume that the order of the numbers is unimportant.
Answer: 1/56 ---- I need help getting this answer. I have tried evrything I can think of. Thank you
Found 2 solutions by Alan3354, galactus:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
In a certain lottery, 3 different numbers between 1 and 8 inclusive are drawn at random. These are the winning number. If you choose 3 different numbers at random between 1 and 8, what is the probability you will match the winning numbers? Assume that the order of the numbers is unimportant.
--------------------
It's not specified, but to get that answer:
No repeats are allowed, ie, 888 is not valid.
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The 1st number is one of 8, then 1 of 7, then 1 of 6.
= 8*7*6
But since is says order doesn't matter, 876 is the same as 678 and 786, etc
That gives each possible number 6 different ways it can be selected.
--> 8*7*6/6
= 8*7
= 1/56
Answer by galactus(183) (Show Source): You can put this solution on YOUR website!
Since order is not important, we use combinations.
C(8,3)=8!/(3!*5!)=56. There are 56 ways to choose 3 items from 8.
Thus, the probability is 1/56.
Another way. Since there is no replacement, we have 3 choices out of 8 for the first number, 2 choices out of 7 for the second, and 1 choice out of 6 for the third.
(3/8)(2/7)(1/6)=1/56
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