If four integers, a, b, c, and d are chosen at random with replacement

Question

If four integers, a, b, c, and d are chosen at random, with replacement, what is the probability that ab + cd is even?

A. 1/4
 
B. 3/8
 
C. 1/2
 
D. 5/8
 
E. 3/4

CaptainOfMySoul · Accepted Answer

Two ways that: ab + cd is even
ab=Odd and cd=Odd (O)
ab=Even and cd=Even (E)

how many ways two arrange ab:
a=O and b=O => ab=O
a=E and b=O => ab=E
a=O and b=E => ab=E
a=E and b=E => ab=E

so ab=E is 3/4 the case and ab=O is 1/4 the case

same structure for cd.

So ab + cd = E = 3/4*3/4 (probability that ab is even multiplied with probability that cd is even) + 1/4*1/4 (probability that ab is odd multiplied with probability that cd is odd) = 5/8

D.

pushpitkc · Answer

We need to know the following   rules   in order to solve this problem:
 Even*Odd = Odd*Even = Even*Even = Even | Odd*Odd = Odd
Even + Odd = Odd + Even = Odd | Odd + Odd = Even + Even = Even

---a-----b-----c-----d------ab+cd--
---E-----E-----E-----E---------E-----
---E-----E-----E-----O---------E-----
---E-----E-----O-----E---------E-----
---E-----E-----O-----O--------O-----
---E-----O-----E-----E---------E-----
---E-----O-----E-----O---------E-----
---E-----O-----O-----E---------E-----
---E-----O-----O-----O---------O-----
---O-----E-----E-----E----------E-----
---O-----E-----E-----O----------E-----
---O-----E-----O-----E----------E-----
---O-----E-----O-----O----------O-----
---O-----O-----E-----E----------O-----
---O-----O-----E-----O----------O-----
---O-----O-----O-----E----------O-----
---O-----O-----O-----O----------E-----

There are a total of 16 possibilities, of which the sum of ab and cd is even in 10 possibilities

Therefore, there are  \frac{10}{16} = \frac{5}{8}  possibilities(Option D) when ab+cd is even

getahead · Answer

Oh wow I could not have thought of 16 possibilities like that... 
My mind went like this 
X=ab and Y=cd 
For Sum of x and Y to be even it's possible in 2 cases 
ODD +Odd is Even 
Even + Even is Even 
And other possibilities are 
Even+Odd= Odd
Odd+ Even= Odd 
So of Total 4 possibilities irrespective of what ab and cd could be the possibility fortheir summation to be even is 2/4 =1/2

Sent from my iPhone using GMAT Club Forum

teaserbae · Answer

pushpitkc   Bunuel 
What's wrong in my thinking ?
 A number can be either odd or even so the probabiblity is half i.e. 1/2

pushpitkc · Answer

teaserbae

As you have clearly mentioned, the probabilty of a number being odd/even is  \frac{1}{2}

Here we are concerned about the probabilities of 4 numbers.
Therefore, the overall probability will have  2^4  or  16  possibilities

Hope this helps you!

Jbbaille · Answer

ab+cd = E so we have:
case 1: Odd + Odd. The only way to express an odd is OO => Probability = 1/16
case 2: Even + Even. 3 ways to express an even number: EE, EO and OE so with the addition, that's 3x3=9 ways, or a probability of 9/16

Case 1 + case 2 = 1/16 + 9/16 = 5/8

draftpunk · Answer

This question is ambiguous. Does "AB" imply multiplication of A x B, or digits in tens & singles places _A_ _B_?

Would GMAT have questions with this degree of ambiguity?

Bunuel

Bunuel · Answer

While I understand what you mean, if ab where two digit integer, then it would have been clearly mentioned. Without this ab can only mean a*b (only multiplication sign can be omitted this way).

agarwalankur · Answer

We can also do it with Combinations:

every no. can be even or odd. Therefore total cases = 2*2*2*2= 16 cases

Now ab + cd can be even when 
1)ab is even and cd is even- ab can be even with three possibilities: O*E, E*O, E*E. Same goes for cd. Therefore 3*3 = 9 cases 
2)ad is odd and cd is odd. ad can be odd with 1 possibility: O*O. Same goes for cd. Therefor 1*1 =1 case

Ans: (9+1)/16 =5/8. Option D

ScottTargetTestPrep · Answer

Solution:

Since odd + odd = even and even + even = even, then either both ab and cd are odd or they are both even.

If both ab and cd are odd, then a, b, c and d have to be all odd. So the probability is ½ x ½ x ½ x ½ = 1/16 (notice that the probability being odd is ½ and the probability being even is also ½).

If both ab and cd are even, then we have:

1) a = even, b = odd, c = even, d = odd

2) a = even, b = odd, c = odd, d = even

3) a = even, b = odd, c = even, d = even

4) a = odd, b = even, c = even, d = odd

5) a = odd, b = even, c = odd, d = even

6) a = odd, b = even, c = even, d = even

7) a = even, b = even, c = even, d = odd

8) a = even, b = even, c = odd, d = even

9) a = even, b = even, c = even, d = even

The probability of any of the cases above is also 1/16. Therefore, the overall probability is 1/16 + 9 x 1/16 = 10/16 = 5/8.

Alternate Solution:

ab + cd is even if both ab and cd is even or both ab and cd is odd. Let’s find P(ab is even) (which equals P(cd is even)) and P(ab is odd) (which equals P(cd is odd)):

P(ab is odd) = P(a is odd and b is odd) = 1/2 x 1/2 = 1/4

P(ab is even) = 1 - P(ab is odd) = 1 - 1/4 = 3/4

Now, the probability that both ab and cd are even is found to be P(both ab and cd are even) = P(ab is even) x P(cd is even) = 3/4 x 3/4 = 9/16. Similarly, P(both ab and cd are odd) = P(ab is odd) x P(cd is odd) = 1/4 x 1/4 = 1/16. Thus, P(ab + cd is even) = P(both ab and cd are even) + P(both ab and cd are odd) = 9/16 + 1/16 = 10/16 = 5/8.

Answer: D

Bambi2021 · Answer

Totally agree. Thought they meant two digit integers ab and cd.

Getting the prompt straight, it gets quite easy:

The Products ab and cd can be even in 3 ways and odd in 1 way, respectively, as products are only odd when two odd numbers are multiplied. Total number of combinations ab + cd = 4*4 = 16. Favourable 10/16.

LaveenaPanchal · Answer

The probability of ab + cd is even is given by

(1) ab = odd and cd = odd i.e. all odd or (2) ab = even (at least one even) and cd = even (atleast one even)

=1/2 * 1/2 *1/2 *1/2 + 3/4 (at least one of a and b even) x 3/4 (at least one of c and d even)

= 1/16 + 9/16 = 10/16 = 5/8

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If four integers, a, b, c, and d are chosen at random with replacement

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