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Re: Hard one ! [#permalink]
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19 Aug 2011, 19:18
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, 09:27
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, 06:26
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, 07:41
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.
Answer: E.



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Re: If x is a randomly chosen integer between 1 and 20 [#permalink]
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28 Aug 2015, 10:57
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 2140 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, 01:48
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. \(P(xy=multiple \ of \ 4)=1(\frac{1}{4}+\frac{1}{8}+\frac{1}{8})=\frac{1}{2}\) Answer: E. ` 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, 07:42
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. Answer: E
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Re: If x is a randomly chosen integer between 1 and 20
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