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

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

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New post 11 Apr 2018, 20:31
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A
B
C
D
E

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Question Stats:

21% (02:43) correct 79% (02:10) wrong based on 182 sessions

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

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New post 23 May 2020, 02:11
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4
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.
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If four integers, a, b, c, and d are chosen at random with replacement  [#permalink]

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New post 11 Apr 2018, 23:05
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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


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

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New post 12 Apr 2018, 06:11
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]

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New post 25 Apr 2018, 06:41
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]

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New post 25 Apr 2018, 08:29
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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New post 22 May 2020, 15:57
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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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New post 24 May 2020, 11:00
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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New post 24 May 2020, 11:10
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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New post 26 May 2020, 12:22
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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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New post 30 May 2020, 13:28
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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Re: If four integers, a, b, c, and d are chosen at random with replacement   [#permalink] 30 May 2020, 13:28

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

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