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12 Aug 2026, 21:51
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
23% (03:34) correct
77%
(03:01)
wrong
based on 31
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\(A\)= {1, 2, 3, 4, 5} and \(B\) = {1, 2, 3, 4, 5, 10}
Four distinct numbers are chosen at random from set \(A\) to form \(A’\), and four distinct numbers are chosen at random from set \(B\) to form \(B’\). What is the probability that the sum of the numbers in \(A’\) is greater than the sum of the numbers in \(B’\)?
(A) \(\frac{1}{15}\)
(B) \(\frac{2}{25}\)
(C) \(\frac{3}{25}\)
(D) \(\frac{2}{15}\)
(E) \(\frac{1}{5}\)
Four distinct numbers are chosen at random from set \(A\) to form \(A’\), and four distinct numbers are chosen at random from set \(B\) to form \(B’\). What is the probability that the sum of the numbers in \(A’\) is greater than the sum of the numbers in \(B’\)?
(A) \(\frac{1}{15}\)
(B) \(\frac{2}{25}\)
(C) \(\frac{3}{25}\)
(D) \(\frac{2}{15}\)
(E) \(\frac{1}{5}\)
13 Aug 2026, 01:23
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In this question, we are given:
A = {1, 2, 3, 4, 5} and B = {1, 2, 3, 4, 5, 10}
We choose 4 numbers from each set, and we want P(Sum of A') > P(Sum of B')
If 10 is selected in B', then the smallest possible sum of B' = 10 + 1 + 2 + 3 = 16, but the largest possible sum of A' = 2+3+4+5 = 14.
Thus, if 10 is selected, success is impossible. Hence, B' must exclude 10.
Now for calculating the probability that B' excludes 10 => There are 6C4 = 15 ways to choose 4 numbers from B. To exclude 10, choose 4 from the other 5 - 5C4 = 5.
Therefore, probability of B' excluding 10 = 5/15 = 1/3
Now, compare the possible sums:
Once 10 is excluded, both A' and B' are choosing 4 numbers from: {1,2,3,4,5}
There are exactly 5 possible 4-number selections:
{1,2,3,4} Sum = 10
{1,2,3,5} Sum = 11
{1,2,4,5} Sum = 12
{1,3,4,5} Sum = 13
{2,3,4,5} Sum = 14
So, each sum is one of 10, 11, 12, 13 and 14.
Count when A' has the greatest sum: 5*5 = 25 possible pairs of sum
We need the A' sum to be strictly greater:
If A' = 10 => 0 possibilities
If A' = 11 => B' = 10 -> 1 possibility
If A' = 12 => B' = 10, 11 -> 2 possibilities
If A' = 13 => B' = 10, 11 and 12 -> 3 possibilities
If A' = 14 => B' = 10, 11, 12 and 13 -> 4 possibilities
Total = 0+1+2+3+4 = 10
Thus, P(A' > B' | B' excludes 10) = 10/25 = 2/5
Combining the probabilities as both are necessary: 1/3 * 2/5 = 2/15
Hence, the correct answer is Option A = 2/15.
Hope this helps!
A = {1, 2, 3, 4, 5} and B = {1, 2, 3, 4, 5, 10}
We choose 4 numbers from each set, and we want P(Sum of A') > P(Sum of B')
If 10 is selected in B', then the smallest possible sum of B' = 10 + 1 + 2 + 3 = 16, but the largest possible sum of A' = 2+3+4+5 = 14.
Thus, if 10 is selected, success is impossible. Hence, B' must exclude 10.
Now for calculating the probability that B' excludes 10 => There are 6C4 = 15 ways to choose 4 numbers from B. To exclude 10, choose 4 from the other 5 - 5C4 = 5.
Therefore, probability of B' excluding 10 = 5/15 = 1/3
Now, compare the possible sums:
Once 10 is excluded, both A' and B' are choosing 4 numbers from: {1,2,3,4,5}
There are exactly 5 possible 4-number selections:
{1,2,3,4} Sum = 10
{1,2,3,5} Sum = 11
{1,2,4,5} Sum = 12
{1,3,4,5} Sum = 13
{2,3,4,5} Sum = 14
So, each sum is one of 10, 11, 12, 13 and 14.
Count when A' has the greatest sum: 5*5 = 25 possible pairs of sum
We need the A' sum to be strictly greater:
If A' = 10 => 0 possibilities
If A' = 11 => B' = 10 -> 1 possibility
If A' = 12 => B' = 10, 11 -> 2 possibilities
If A' = 13 => B' = 10, 11 and 12 -> 3 possibilities
If A' = 14 => B' = 10, 11, 12 and 13 -> 4 possibilities
Total = 0+1+2+3+4 = 10
Thus, P(A' > B' | B' excludes 10) = 10/25 = 2/5
Combining the probabilities as both are necessary: 1/3 * 2/5 = 2/15
Hence, the correct answer is Option A = 2/15.
Hope this helps!
13 Aug 2026, 23:00
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If B’ doesn’t include 10, what’s the probability that the sum of the numbers in A’ will be equal to the sum of the numbers in B’?
