Slips of paper with the numbers 1-4 are placed in a hat. Two slips of paper are drawn WITHOUT REPLACEMENT. Chance that two is at least 5?

Slips of paper with the numbers 1,2,3 and 4 are placed in a hat. Two slips of paper are drawn WITHOUT REPLACEMENT (which means that the first slip is drawn, the number on it is noted, then the slip is removed).

What is the probability the sum of the values is at least 5?

I edited the question a bit, but thats basically what it is asking. It was originally a 2 part question, but i solved the other part, not entirely sure how to solve this portion.

Any advice would be appreciated, thanks!

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Think of the sample space: the set of all possible outcomes.

1,2 1,3 1,4

2,1, 2,3 2,4

3,1 3,2 3,4

4,1 4,2, 4,3

So there are 12 outcomes possible. Now circle all of the sums greater than or equal to five...you should find that there are 8.

8/12 = 2/3.

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