GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 07 Jul 2020, 15:37 ### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # If four integers, a, b, c, and d are chosen at random with replacement

Author Message
TAGS:

### Hide Tags

Math Expert V
Joined: 02 Sep 2009
Posts: 65062
If four integers, a, b, c, and d are chosen at random with replacement  [#permalink]

### Show Tags

22 00:00

Difficulty:   95% (hard)

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

### HideShow timer Statistics

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

_________________
Intern  B
Joined: 14 Apr 2019
Posts: 9
Location: Germany
Concentration: Entrepreneurship, Technology
WE: Engineering (Energy and Utilities)
Re: If four integers, a, b, c, and d are chosen at random with replacement  [#permalink]

### Show Tags

4
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.
##### General Discussion
Senior PS Moderator V
Joined: 26 Feb 2016
Posts: 3249
Location: India
GPA: 3.12
If four integers, a, b, c, and d are chosen at random with replacement  [#permalink]

### Show Tags

1
3
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
_________________
You've got what it takes, but it will take everything you've got
Intern  B
Joined: 16 Mar 2018
Posts: 46
Location: United Kingdom
GMAT 1: 650 Q45 V35
Re: If four integers, a, b, c, and d are chosen at random with replacement  [#permalink]

### Show Tags

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
Manager  S
Joined: 24 Mar 2018
Posts: 236
If four integers, a, b, c, and d are chosen at random with replacement  [#permalink]

### Show Tags

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
Senior PS Moderator V
Joined: 26 Feb 2016
Posts: 3249
Location: India
GPA: 3.12
If four integers, a, b, c, and d are chosen at random with replacement  [#permalink]

### Show Tags

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!
_________________
You've got what it takes, but it will take everything you've got
Intern  B
Joined: 13 Feb 2020
Posts: 14
Re: If four integers, a, b, c, and d are chosen at random with replacement  [#permalink]

### Show Tags

1
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
Manager  S
Joined: 28 Jan 2017
Posts: 52
Re: If four integers, a, b, c, and d are chosen at random with replacement  [#permalink]

### Show Tags

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
Math Expert V
Joined: 02 Sep 2009
Posts: 65062
Re: If four integers, a, b, c, and d are chosen at random with replacement  [#permalink]

### Show Tags

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).
_________________
Intern  B
Joined: 02 Feb 2019
Posts: 7
Re: If four integers, a, b, c, and d are chosen at random with replacement  [#permalink]

### Show Tags

1
1
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
Target Test Prep Representative V
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 11043
Location: United States (CA)
Re: If four integers, a, b, c, and d are chosen at random with replacement  [#permalink]

### Show Tags

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.

_________________

# Scott Woodbury-Stewart

Founder and CEO

Scott@TargetTestPrep.com

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews 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   