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Re: If four integers, a, b, c, and d are chosen at random with replacement [#permalink]
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


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If four integers, a, b, c, and d are chosen at random with replacement [#permalink]
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
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If four integers, a, b, c, and d are chosen at random with replacement [#permalink]
teaserbae wrote:
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


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!
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Re: If four integers, a, b, c, and d are chosen at random with replacement [#permalink]
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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
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Re: If four integers, a, b, c, and d are chosen at random with replacement [#permalink]
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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?

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Re: If four integers, a, b, c, and d are chosen at random with replacement [#permalink]
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costcosized wrote:
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


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).
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Re: If four integers, a, b, c, and d are chosen at random with replacement [#permalink]
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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
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Re: If four integers, a, b, c, and d are chosen at random with replacement [#permalink]
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Bunuel wrote:
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


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
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If four integers, a, b, c, and d are chosen at random with replacement [#permalink]
draftpunk wrote:
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


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.
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Re: If four integers, a, b, c, and d are chosen at random with replacement [#permalink]
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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Re: If four integers, a, b, c, and d are chosen at random with replacement [#permalink]
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Re: If four integers, a, b, c, and d are chosen at random with replacement [#permalink]
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