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What is the probability that the sum of two different number
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06 May 2019, 12:04
06 May 2019, 12:04
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Expert Reply
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Question Stats:
61% (01:20) correct
38% (00:58) wrong based on 36 sessions
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What is the probability that the sum of two different numbers randomly picked (without replacement) from the set S = {1, 2, 3, 4} is 5?
(A) \(\frac{1}{5}\)
(B) \(\frac{3}{16}\)
(C) \(\frac{1}{4}\)
(D) \(\frac{1}{3}\)
(E) \(\frac{1}{2}\)
(A) \(\frac{1}{5}\)
(B) \(\frac{3}{16}\)
(C) \(\frac{1}{4}\)
(D) \(\frac{1}{3}\)
(E) \(\frac{1}{2}\)
Re: What is the probability that the sum of two different number
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12 Oct 2019, 05:16
12 Oct 2019, 05:16
2
1
To get 5 we need 2,3 or 1,4.
We can pick any number at first so 4/4.
For the second pick, however, there is just one available number: 1/3
Hence, 4/4*1/3 = 1/3.
Answer is D.
We can pick any number at first so 4/4.
For the second pick, however, there is just one available number: 1/3
Hence, 4/4*1/3 = 1/3.
Answer is D.
Re: What is the probability that the sum of two different number
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12 Oct 2019, 05:28
12 Oct 2019, 05:28
2
Another way of solving it is by counting.
Since the number of elements in the set is small we can write down the possible combinations:
1,2 1,3 1,4 2,3 2,4 3,4 and vice versa, hence 12.
Of the couples above the ones respecting the condition are 1,4 2,3 and 4,1 3,2, hence 4.
4 out of 12 is 1/3.
Since the number of elements in the set is small we can write down the possible combinations:
1,2 1,3 1,4 2,3 2,4 3,4 and vice versa, hence 12.
Of the couples above the ones respecting the condition are 1,4 2,3 and 4,1 3,2, hence 4.
4 out of 12 is 1/3.
