If x is a randomly chosen integer between 1 and 20, inclusive

Question

If x is a randomly chosen integer between 1 and 20, inclusive, and y is a randomly chosen integer between 21 and 40, inclusive, what is the probability that xy is a multiple of 4?

(A) 1/4
(B) 1/3
(C) 3/8
(D) 7/16
(E) 1/2

Bunuel · Accepted Answer

If x is a randomly chosen integer between 1 and 20, inclusive, and y is a randomly chosen integer between 21 and 40, inclusive, what is the probability that xy is a multiple of 4?

(A) 1/4
(B) 1/3
(C) 3/8
(D) 7/16
(E) 1/2

Let's find the probability of an opposite event and subtract this value from 1.

There are three cases xy NOT to be a multiple of 4:

A. both x and y are odd --> 1/2*1/2=1/4;
B. x is odd and y is even but not multiple of 4 --> 1/2*1/4=1/8;
C. y is odd and x is even but not multiple of 4 --> 1/2*1/4=1/8.

P(xy=multiple \ of \ 4)=1-(\frac{1}{4}+\frac{1}{8}+\frac{1}{8})=\frac{1}{2}

Answer: E.

brau0300 · Answer

There are two ways that I see of getting a multiple of 4

1) x and/or y is a multiple of 4
2) x is a multiple of 2 and y is a multiple of 2

for 1 we have

P(option 1) = (5/20)(20/20) + (5/20)(20/20) - (5/20)(5/20) = 1/2 - 1/16

since 5 of the numbers in each range are divisible by 4, and if either x or y is selected to be divisible by 4 the selection of the other variable does not matter. Notice the last subtraction term which corrects the fact that we have counted the possibility of both being divisible by 4 twice (e.g. P(A union B) = P(A) + P(B) - P(A and B)).

Notice at this point that the probability of option one is already 7/16, the second largest available option. If we can come up with any other possibilities for getting xy divisble by 4 we must go with option E.

Of course this is the case because if we select a multiple of 2 that is not a multiple of 4 from both ranges of numbers, the product is divisible by 4.

We can formally compute this as:

P(option 2) = (5/20) * (5/20) = 1/16

since we do not wish to double count possibilities that fall into category 1, we only count the probability of selecting even numbers from each set that are not divisible by 4 (e.g. the other 5 even integers in each set).

Total probability is then P(option 1) + P(option 2) = 1/2

rxs0005 · Answer

the way i approached it

Case 1 x is a multiple of 4 5/20 ( 5 multiples of 4 in that range ) = 1/4

it does not matter what y is XY will be a multiple of 4

Case 2 same as case 1 execpt it does not matter what x is so 1/4

total P = 1/4 + 1/4 = 1/2

No1EzPerfect · Answer

May be I am wrong but I think this is not a full proof method.
Lets say x=6 (not multiple of 4) and y=22 (not multiple of 4)
but xy = 132 (which is multiple of 4).

Although you got the answer but cannot be justified by the theory.

shrouded1 · Answer

It is incorrect, the answer is only correct because of luck

x is multiple of 4 and y is odd : 1/4* (1/2) = 1/8
y is multiple of 4 and x is odd : 1/4* (1/2) = 1/8
x and y are both even : 1/2*1/2 = 1/4

So net probability = 1/2

john2roll2 · Answer

If x is a randomly chosen integer between 1 and 20, inclusive, and y is a randomly chosen integer between 21 and 40, inclusive, what is the probability that xy is a multiple of 4?

(A) 1/4
(B) 1/3
(C) 3/8
(D) 7/16
(E) 1/2

The solution is given below. However i want to know to solve it by not talking the "  neither  " approach.

Answer:
There are two ways xy can be a multiple of 4. First, if either x or y (or both) is a mulitple of 4, it doesn't matter what the other number is: a multiple of 4 times anything is a mulitple of 4. Second, if neither of the numbers are multiples of 4, but both are even (for instance, 2 and 22), the product will be a multiple of 4. We need to find the probability of each of those possibilities.

To find the probability that either x, y, or both is a multiple of 4, it's easiest to find the probability that NEITHER are multiples of 4. The probability that x is NOT a multiple of 4 is 3/4 (1/4 of numbers are multiples of 4), and the probability that y is NOT a multiple of 4 is also 3/4. Thus, the probability that NEITHER is a multiple of 4 is (3/4)(3/4) = 9/16. Thus, the probability that one or both of the numbers is a multiple of 4 is 1 - 9/16 = 7/16.

That leaves us to solve for the other possibility: that both numbers are even but not multiples of 4. In any sequence of 4 consecutive integers, one of the 4 will be an even number that is not a multiple of 4. Thus, 1/4 of the numbers between 1 and 20 (or 21 and 40) is an even non-multiple of 4. The probability that BOTH numbers have these characteristics is (1/4)(1/4) = 1/16.

The probability that the product is a multiple of 4, then, is the sum of our two probabilities: 
7/16 + 1/16 = 8/16 = 1/2

pkmme · Answer

There are 5 multiples of 4 in the range from 1 to 20 and 21 to 40. For xy to be divisible by 4, we can pick up one of the 5 multiples from one series and any number from the other series.

P = (5c1. 20c1 + 5c1.20c1)/ 20c1.20c1 
 = 200/400 = 1/2

DeeptiM · Answer

There are three cases xy NOT to be a multiple of 4:
A. both x and y are odd --> 1/2*1/2=1/4;
B. x is odd and y is even but not multiple of 4 --> 1/2*1/4=1/8;
C. y is odd and x is even but not multiple of 4 --> 1/2*1/4=1/8.

Well i seem to be doing something wrong for sure for not arriving at 1/2*1/4 equation..

Can someone pls help with the working..

KarishmaB · Answer

You are right. The three cases are as given by you. Now complete the solution.
The total probability that xy is not a multiple of 4 is 1/4 + 1/8 + 1/8 = 1/2
So the probability that xy is a multiple of 4 is 1 - 1/2 = 1/2

mau5 · Answer

Don't think this is needed after Bunuel's explanations, however, would like to try a different approach.

Imagine the question as:

Find the probability that xy is a multiple of 4, where x and y are both randomly chosen integers between   and   both inclusive. This is relatively easier to calculate. 4 cases, where 4*5,4*6,4*7 and 4*8 will be possible, along with 1*8,2*6 and 2*8,3*8  	o \frac{8}{16} = \frac{1}{2}

Now , imagine the same question reads : Find the probability that xy is a multiple of 4, where x and y are both randomly chosen integers between  ,both inclusive.

Will any of these changes make any difference?NO. It is so because the no. of multiples of 4 in   and   are the same,i.e. 5. and this is the only governing principle in this problem. Thus, any asnwer for the given probability other than \frac{1}{2}  would indicate that the no of multiples of 4 in   and   are not the same.

AKG1593 · Answer

Option E.
Total no. of outcomes= 400 
No. of favourable outcomes:
When  x=1  We can have  5  different values of  y(24,28,32,36,40)  giving xy some multiple of  4 .
When  x=2  we can have  10  different values of  y(all even nos. from 21 to 40)  giving xy=some multiple of  4 .
When  x=3  we can have  5  different values of  y(24,28,32,36,40)  giving xy some multiple of  4 .
And when  x=4  we can have 20 different values of  y(from 21 to 40)  giving xy some multiple of  4 .

Likewise this can be repeated 5 times.
So total no. of favourable outcomes= 40*5=200

Probability=200/400=1/2

ScottTargetTestPrep · Answer

We see that xy is a multiple of 4 if both x and y are multiples of 2 (but not multiples of 4), OR x is a multiple of 4, OR y is a multiple of 4, or . Let’s look at the three cases.
Case 1: Both x and y are multiples of 2 but neither is a multiple of 4

Thus x could be 2, 6, 10, 14, or 18 and y could be 22, 26, 30, 34, or 38. We see that there are 5 choices for x and 5 choices for y; thus, the number of integers xy that are a multiple of 4 is 5 x 5 = 25.

Case 2: x is a multiple of 4

If x is a multiple of 4 (i.e., x could be 4, 8, 12, 16, or 20), then y could be any integer. Thus, the number of integers xy that are a multiple of 4 is 5 x 20 = 100.

Case 3: y is a multiple of 4

If y is a multiple of 4 (i.e., y could be 24, 28, 32, 36, or 40), then x could be any integer. Thus, the number of integers xy that are a multiple of 4 is 20 x 5 = 100.

However, there are overlaps in cases 2 and 3. For example, x = 4 and y = 24 is an overlap in both cases. We need to subtract the overlaps. The overlaps occur when both x and y are multiples of 4. Since there are 5 choices for x being a multiple of 4 and another 5 choices for y being a multiple of 4, the total number of overlaps is 5 x 5 = 25.

Thus, the total number of integers xy that are a multiple of 4 is 25 + 100 + 100 - 25 = 200. Since the total number of integers xy that can be formed is 20 x 20 = 400, the probability that the product xy is a multiple of 4 is 200/400 = 1/2.

Answer: E

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If x is a randomly chosen integer between 1 and 20, inclusive

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