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If x is a randomly chosen integer between 1 and 20

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Veritas Prep GMAT Instructor
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Re: Hard one !  [#permalink]

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19 Aug 2011, 18:18
1
DeeptiM wrote:
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..

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
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Re: If x is a randomly chosen integer between 1 and 20  [#permalink]

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07 Jul 2013, 08:27
nusmavrik wrote:
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) ½

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 [1,4] and [5,8] 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 $$\to \frac{8}{16} = \frac{1}{2}$$

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 [1,4],both inclusive.

Will any of these changes make any difference?NO. It is so because the no. of multiples of 4 in [1,20] and [21,40] 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 [1,20] and [21,40] are not the same.
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Re: If x is a randomly chosen integer between 1 and 20  [#permalink]

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29 Mar 2014, 05:26
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=$$40*5=200$$

$$Probability=200/400=1/2$$
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Re: If x is a randomly chosen integer between 1 and 20  [#permalink]

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08 Sep 2014, 06:41
This is actually a relatively simple probability question.

We have 3 cases that satisfy "xy is a multiple of 4":

Case 1) X is a multiple of 4 and Y is not
Case 2) X is not a multiple of 4 and Y is
Case 3) X and Y are BOTH multiples of 4
Case 4) X and Y are BOTH multiples of 2 but NOT multiples of 4.

Notice that these cases are all mutually exclusive and thus we don't have to consider overlaps.

for set X, there are 5 numbers that are multiples of 4: 4, 8, 12, 16, 20; P(x is multiple of 4) = 5/20

for set Y, there are 5 numbers that are multiples of 4: 24, 28, 32, 36, 40; P(y is multiple of 4) = 5/20.

For set X, there are 5 numbers that are multiples of 2 that are not multiples of 4: 2,6,10, 14, 18. P(x is multiple of 2 and not 4) = 5/20

For set Y, there are 5 numbers that are multiples of 2 that are not multiples of 4: 22, 26, 30, 34, 38. P(x is multiple of 2 and not 4) = 5/20.

Now, just calculate the probabilities for all of the cases.

1) (5/20)(15/20) = 75/400
2) (15/20)(5/20) = 75/400
3) (5/20)(5/20) = 25/400
4) (5/20)(5/20) = 25/400

Adding these all together, we have 150+50 = 200/400 = 1/2.

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Re: If x is a randomly chosen integer between 1 and 20  [#permalink]

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28 Aug 2015, 09:57
TOOK 3 , Brunel approach is the fastest way i guess ; why didnt i think of that ;

total combinations =400 ;

when x is 1 3 5 7 9 11 13 15 17 or 19 then y should be 24 28 32 36 40

10C1 * 5C1 = 50 ;

when x is 2 6 10 14 or 18 then y can be 22 24 26 28 30 32 34 36 38 40

5C1 * 10C1 = 50 ;

when x is 4 8 12 16 20 , then y can be anything from 21-40

5C1 * 20C1 = 100 ;

TOTAL = 50 + 50 + 100 = 200

PROB = 200/400 = 1/2

i am not sure this is rght way though
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Re: If x is a randomly chosen integer between 1 and 20  [#permalink]

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18 Jan 2018, 00:48
Bunuel wrote:
nusmavrik wrote:
Please use the timer and let us know the time. The quest for the fastest way !

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) ½

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.

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

`

Bunuel
hi

brilliant solution! +1
do you think, however, the below is okay?

x is a multiple of "4" AND y is anything OR y is a multiple of "4" AND x is anything
which implies

(1/4) * 1 + (1/4) * 1

= 1/2

???

thanks in advance, man
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Re: If x is a randomly chosen integer between 1 and 20  [#permalink]

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20 Jan 2018, 06:42
1
nusmavrik wrote:
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) ½

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.

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Re: If x is a randomly chosen integer between 1 and 20 &nbs [#permalink] 20 Jan 2018, 06:42

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