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11 Aug 2026, 23:39
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If \(v,w,x,y,z\) are the numbers showing up on five successive rolls of a fair six-sided die, what is the probability that \(vwx+yz\) is even?
(A) \(\frac{5}{8}\)
(B) \(\frac{11}{16}\)
(C) \(\frac{3}{4}\)
(D) \(\frac{13}{16}\)
(E) \(\frac{7}{8}\)
(A) \(\frac{5}{8}\)
(B) \(\frac{11}{16}\)
(C) \(\frac{3}{4}\)
(D) \(\frac{13}{16}\)
(E) \(\frac{7}{8}\)
13 Aug 2026, 00:58
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Since we are talking about the outcomes of a fair six-sided die in this question, v, w, x, y and z can be anything in the range 1 to 6 (inclusive).
Thus,
Probability (Odd Number) = 1/2 (1,3,5 out of 1,2,3,4,5 and 6)
Probability (Even Number) = 1/2 (2,4,6 out of 1,2,3,4,5 and 6)
Now, the main task is to ensure that the sum of vwx + yz is even. The sum of two numbers can be even only in two possible cases:
Case 1 - Both numbers are odd
Case 2 - Both numbers are even
Let's try to think about both the cases one-by-one.
Case 1 - Both numbers are odd, i.e., vwx and yz both are odd.
For vwx to be odd, we would need to ensure that all of the three - v, w and z are odd.
Hence, the probability of vwx being odd = (1/2)^3 and the probability of yz to be odd = (1/2)^2
Since we want both vwx and yz to be odd, probability becomes (1/2)^5 = 1/32
Case 2 - Both numbers are even, i.e., vwx and yz both are even.
This can be easily calculated by using Case 1. Since there can be only two possible outcomes - even and odd, probability of vwx being even = 1 - (1/2)^3 = 7/8 and yz being even = 1 - (1/2)^2 = 3/4
Hence, the total probability for this case = 7/8 * 3/4 = 21/32
Thus, the final answer becomes 1/32 + 21/32 (as we would want either of the cases to be true for the sum to be even) = 22/32 = 11/16
Thus, the correct answer is Option B - 11/16.
Hope this helps!
Thus,
Probability (Odd Number) = 1/2 (1,3,5 out of 1,2,3,4,5 and 6)
Probability (Even Number) = 1/2 (2,4,6 out of 1,2,3,4,5 and 6)
Now, the main task is to ensure that the sum of vwx + yz is even. The sum of two numbers can be even only in two possible cases:
Case 1 - Both numbers are odd
Case 2 - Both numbers are even
Let's try to think about both the cases one-by-one.
Case 1 - Both numbers are odd, i.e., vwx and yz both are odd.
For vwx to be odd, we would need to ensure that all of the three - v, w and z are odd.
Hence, the probability of vwx being odd = (1/2)^3 and the probability of yz to be odd = (1/2)^2
Since we want both vwx and yz to be odd, probability becomes (1/2)^5 = 1/32
Case 2 - Both numbers are even, i.e., vwx and yz both are even.
This can be easily calculated by using Case 1. Since there can be only two possible outcomes - even and odd, probability of vwx being even = 1 - (1/2)^3 = 7/8 and yz being even = 1 - (1/2)^2 = 3/4
Hence, the total probability for this case = 7/8 * 3/4 = 21/32
Thus, the final answer becomes 1/32 + 21/32 (as we would want either of the cases to be true for the sum to be even) = 22/32 = 11/16
Thus, the correct answer is Option B - 11/16.
Hope this helps!
General Discussion
13 Aug 2026, 00:31
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What must be true of two integers for their sum to be even ?
