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# Two integers between 1 and 100, inclusive, each randomly and independe

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Math Expert
Joined: 02 Sep 2009
Posts: 43312

Kudos [?]: 139308 [0], given: 12783

Two integers between 1 and 100, inclusive, each randomly and independe [#permalink]

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24 Apr 2015, 02:03
Expert's post
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Difficulty:

85% (hard)

Question Stats:

48% (01:41) correct 52% (02:08) wrong based on 103 sessions

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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.
[Reveal] Spoiler: OA

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Kudos [?]: 139308 [0], given: 12783

Manager
Joined: 03 Sep 2014
Posts: 75

Kudos [?]: 51 [3], given: 89

Concentration: Marketing, Healthcare
Re: Two integers between 1 and 100, inclusive, each randomly and independe [#permalink]

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24 Apr 2015, 10:57
3
KUDOS
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 = $$(1 - 3/8) = 5/8$$

Kudos [?]: 51 [3], given: 89

Current Student
Joined: 29 Mar 2015
Posts: 77

Kudos [?]: 26 [1], given: 21

Concentration: Strategy, Operations
GMAT 1: 700 Q51 V33
WE: Research (Other)
Re: Two integers between 1 and 100, inclusive, each randomly and independe [#permalink]

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24 Apr 2015, 13:47
1
KUDOS
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 1-100 total 50 even, 50 odd so p(even) = p(odd) = 1/2
Cases: even+even, even*even, odd+odd, odd*even, even*odd
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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Kudos [?]: 26 [1], given: 21

Director
Joined: 07 Aug 2011
Posts: 579

Kudos [?]: 606 [3], given: 75

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Re: Two integers between 1 and 100, inclusive, each randomly and independe [#permalink]

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24 Apr 2015, 19:09
3
KUDOS
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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Kudos [?]: 606 [3], given: 75

Manager
Joined: 24 Jan 2015
Posts: 67

Kudos [?]: 154 [1], given: 1

GMAT 1: 590 Q39 V31
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Re: Two integers between 1 and 100, inclusive, each randomly and independe [#permalink]

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25 Apr 2015, 05:02
1
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1
This post was
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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 = $$[\frac{1}{2} * \frac{1}{2}] + [\frac{1}{2} * \frac{3}{4}]$$ = $$\frac{1}{4} + \frac{3}{8}$$ = $$\frac{5}{8}$$

Kudos [?]: 154 [1], given: 1

Manager
Joined: 03 May 2013
Posts: 74

Kudos [?]: 15 [1], given: 105

Two integers between 1 and 100, inclusive, each randomly and independe [#permalink]

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26 Sep 2015, 23:06
1
KUDOS
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]

Kudos [?]: 15 [1], given: 105

Math Expert
Joined: 02 Aug 2009
Posts: 5520

Kudos [?]: 6417 [1], given: 122

Re: Two integers between 1 and 100, inclusive, each randomly and independe [#permalink]

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26 Sep 2015, 23:19
1
KUDOS
Expert's post
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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Kudos [?]: 6417 [1], given: 122

Manager
Joined: 03 May 2013
Posts: 74

Kudos [?]: 15 [0], given: 105

Re: Two integers between 1 and 100, inclusive, each randomly and independe [#permalink]

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26 Sep 2015, 23:44
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,

Kudos [?]: 15 [0], given: 105

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Re: Two integers between 1 and 100, inclusive, each randomly and independe [#permalink]

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09 Dec 2017, 05:34
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Re: Two integers between 1 and 100, inclusive, each randomly and independe   [#permalink] 09 Dec 2017, 05:34
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