If v,w,x,y,z are the numbers showing up on five succe

Question

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}

shivani1351 · Accepted Answer

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!

kevincan · Answer

What must be true of two integers for their sum to be even ?

kevincan · Answer

Wonferfully pedagogic explanation ! Thanks for your generosity in taking the time to write it !

shivani1351 · Answer

Thank you kevincan , means a lot to me

prathameshsingh · Answer

How are we assuming that v w x y and z can be identical because in the odd case.

Lets choose v as odd ( Prob of 1/2) then when choosing w i have ( 2 Odd from a total pool of 5 - 2/5), unless i assume a case where v and w are same.

Please help resolve this doubt

kevincan · Answer

Rolls are independent, so two or more successive rolls may well produce the same result

EarlyHope · Answer

For (v*w*x) to be even only one has to nececarrily be even the other numbers can be even or odd but for it to be odd all of them have to be odd so simply subtracting the 2 can't give the correct resluts

If v,w,x,y,z are the numbers showing up on five succe

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If v,w,x,y,z are the numbers showing up on five succe

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