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Two integers between 1 and 100, inclusive, each randomly and independe [#permalink]
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24 Apr 2015, 03:03
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48% (01:37) correct 52% (02:09) wrong based on 108 sessions
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Re: Two integers between 1 and 100, inclusive, each randomly and independe [#permalink]
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24 Apr 2015, 11:57
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Bunuel wrote: Two integers between 1 and 100, inclusive, each randomly and independently chosen, are either added or multiplied, with an equal chance of either operation. What is the probability that the result is even?
(A) 1/3 (B) 1/2 (C) 5/8 (D) 2/3 (E) 7/8
Kudos for a correct solution. Probability of getting even = 1  Probability of getting Odd Getting Odd outcome = Odd*Odd or Odd + Even or Even + Odd Thus, Probability of picking Odd, Odd and multiplication = \(1/2*1/2*1/2 = 1/8\) Probability of picking Odd, Even and addition = \(1/2*1/2*1/2 = 1/8\) Probability of picking Even, Odd and addition = \(1/2*1/2*1/2 = 1/8\) Thus, Probability of getting an Odd number = \(1/8 + 1/8 + 1/8 = 3/8\) And Probability of getting an Even number = \((1  3/8) = 5/8\) Hence, answer is C



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Re: Two integers between 1 and 100, inclusive, each randomly and independe [#permalink]
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24 Apr 2015, 14:47
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Bunuel wrote: Two integers between 1 and 100, inclusive, each randomly and independently chosen, are either added or multiplied, with an equal chance of either operation. What is the probability that the result is even?
(A) 1/3 (B) 1/2 (C) 5/8 (D) 2/3 (E) 7/8
Kudos for a correct solution. Nos from 1100 total 50 even, 50 odd so p(even) = p(odd) = 1/2 Cases: even+even, even*even, odd+odd, odd*even, even*odd P(multiplication, addition) = 1/2 hence for every case Probability = 1/8 >(P of add / multiply * p of 1st no. * P of picking second no.) , total 5 cases so 5/8. (C)
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Re: Two integers between 1 and 100, inclusive, each randomly and independe [#permalink]
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24 Apr 2015, 20:09
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1/2×1/2×1/2 = P (+, odd, odd) 1/2×1/2×1/2 = P (+, even, even) 1/2×1/2×1/2 = P (×, odd, even) 1/2×1/2×1/2 = P (×, even, odd) 1/2×1/2×1/2 = P (×, even, even) 5×1/2×1/2×1/2= 5/8
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Re: Two integers between 1 and 100, inclusive, each randomly and independe [#permalink]
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25 Apr 2015, 06:02
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Total number of integers = 100 Number of Even integers = 50 Number of Odd integers = 50
Probability of selecting an odd integer = \(\frac{50}{100}\) = \(\frac{1}{2}\) Probability of selecting an even integer = \(\frac{50}{100}\) = \(\frac{1}{2}\)
From number properties, we know that for addition of two integers to be even => Odd + Odd = Even or Even + Even = Even
Using the above cases, the probability that the addition of two integers is even (Pa)= \([\frac{1}{2} * \frac{1}{2}] + [\frac{1}{2} * \frac{1}{2}]\) = \(\frac{1}{4} + \frac{1}{4}\) = \(\frac{1}{2}\)
Similarly , from number properties, we know that for multiplication of two integers to be even => Even * Odd = Even or Odd * Even = Even or Even * Even = Even
Using the above cases, the probability that the multiplication of two integers is even (Pm)= \([\frac{1}{2} * \frac{1}{2}] + [\frac{1}{2} * \frac{1}{2}] + [\frac{1}{2} * \frac{1}{2}]\) = \(\frac{1}{4} + \frac{1}{4} + \frac{1}{4}\) = \(\frac{3}{4}\)
From the question stem , we know that both addition and multiplication have an equal chance i.e. \(\frac{1}{2}\) * Pa and \(\frac{1}{2}\) * Pm
Therefore, the total Probability for the result to be even = \(\frac{1}{2}\) * Pa + \(\frac{1}{2}\) * Pm = \([\frac{1}{2} * \frac{1}{2}] + [\frac{1}{2} * \frac{3}{4}]\) = \(\frac{1}{4} + \frac{3}{8}\) = \(\frac{5}{8}\)
Answer is (C)



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Two integers between 1 and 100, inclusive, each randomly and independe [#permalink]
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27 Sep 2015, 00:06
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Bunuel wrote: Two integers between 1 and 100, inclusive, each randomly and independently chosen, are either added or multiplied, with an equal chance of either operation. What is the probability that the result is even?
(A) 1/3 (B) 1/2 (C) 5/8 (D) 2/3 (E) 7/8
Kudos for a correct solution. Hi experts , please shed some light , first of all by picking randomly , I think it is talking abt without replacement, which means P of selecting first even no  50/100 then second 49/99(not again 50/100) second; why double counted sets are not subtracted , for example picking 40 and 42 give even result whether the opreation may be app or multiply , hence it should counted once only and in every solution it is counted twice. in my opinion the sol should be(considering with replacement) [1/2 {ee(even,even)+oo(odd,odd)} + 1/2 (eo+oe+ee)  ee]



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Re: Two integers between 1 and 100, inclusive, each randomly and independe [#permalink]
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27 Sep 2015, 00:19
vipulgoel wrote: Bunuel wrote: Two integers between 1 and 100, inclusive, each randomly and independently chosen, are either added or multiplied, with an equal chance of either operation. What is the probability that the result is even?
(A) 1/3 (B) 1/2 (C) 5/8 (D) 2/3 (E) 7/8
Kudos for a correct solution. Hi experts , please shed some light , first of all by picking randomly , I think it is talking abt without replacement, which means P of selecting first even no  50/100 then second 49/99(not again 50/100) second; why double counted sets are not subtracted , for example picking 40 and 42 give even result whether the opreation may be app or multiply , hence it should counted once only and in every solution it is counted twice. in my opinion the sol should be(considering with replacement) [1/2 {ee(even,even)+oo(odd,odd)} + 1/2 (eo+oe+ee)  ee] hi, 1) concerning your first doubt , the two numbers are picked up independent of each other so the same number can be picked up twice.. 2) secondly each multiplication operation is different from and independent of add operation and possiblity of add or multiplication is 1/2 so you have to take event separately even if the two operations lead to same outcome hope it helped
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Re: Two integers between 1 and 100, inclusive, each randomly and independe [#permalink]
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27 Sep 2015, 00:44
chetan2u wrote: vipulgoel wrote: Bunuel wrote: Two integers between 1 and 100, inclusive, each randomly and independently chosen, are either added or multiplied, with an equal chance of either operation. What is the probability that the result is even?
(A) 1/3 (B) 1/2 (C) 5/8 (D) 2/3 (E) 7/8
Kudos for a correct solution. Hi experts , please shed some light , first of all by picking randomly , I think it is talking abt without replacement, which means P of selecting first even no  50/100 then second 49/99(not again 50/100) second; why double counted sets are not subtracted , for example picking 40 and 42 give even result whether the opreation may be app or multiply , hence it should counted once only and in every solution it is counted twice. in my opinion the sol should be(considering with replacement) [1/2 {ee(even,even)+oo(odd,odd)} + 1/2 (eo+oe+ee)  ee] hi, 1) concerning your first doubt , the two numbers are picked up independent of each other so the same number can be picked up twice.. 2) secondly each multiplication operation is different from and independent of add operation and possiblity of add or multiplication is 1/2 so you have to take event separately even if the two operations lead to same outcome hope it helped Thanks for the prompt response about #2 i m still skeptical , i understand operations are different , but if outcome is coming out same twice then why we are not subtracting , ne how we are concerned with final outcome,



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Re: Two integers between 1 and 100, inclusive, each randomly and independe [#permalink]
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09 Dec 2017, 06:34
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Re: Two integers between 1 and 100, inclusive, each randomly and independe
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