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

Question

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.

(A) 1/3 
(B) 1/2 
(C) 5/8 
(D) 2/3 
(E) 7/8

King407 · Accepted Answer

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

AVRonaldo · Answer

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}{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}{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}{4} + \frac{3}{8}  =  \frac{5}{8}

Answer is (C)

M3sSiaH · Answer

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 
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)

Lucky2783 · Answer

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

vipulgoel · Answer

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)

chetan2u · Answer

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

vipulgoel · Answer

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,

Manas1212 · Answer

Hi  VeritasKarishma  ,  chetan2u

I am unable to understand how can we have 8 possible combinations ?

(Even,odd) (Odd,odd) or (Even,Even)

Considering addition or Multiplication , total combination is 6 and no of cases in which we will get even is 6, making it 2/3.

Even we are picking up the same no again, aren't they covered as a part of (e,e) or (o,o) ?
Please help.

chetan2u · Answer

What you are missing out is (odd,even) is different from (even,odd)..
So (Odd,even), (Even,odd) (Odd,odd) or (Even,Even) are 4 different possibilities..
Both operations have equal chance so 1/2..
Addition ..
Out of 4 possibilities, 2 will result in odd and 2 in even..
So probability is (1/2)*(2/4)=1/4
Multiplication..
Out of possibilities, only 1 will result in odd when both are odd.
So probability is (1/2)*(3/4)=3/8

Total probability is 1/4+3/8=5/8

C

Krishnahelps · Answer

I tried doing this way- 1- (1/2*(50c2+2!(50c1*50c1))/100c2)

50c2- Choosing from the set of odd numbers
50c1- Choosing each from the set of odd numbers and even numbers

Where am I going wrong? Please help

theint481 · Answer

I am confusing multiplication of 1/2 in 3 times for picking two integers. 
Thus means, the probability of (Odd + Odd) = 1/2 * 1/2 = 1/4 (not 1/8).
Please kindly answer me if you get the answer. Thank you..

Bunuel · Answer

Check the solution below, hope it answers your question.

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.

(A) 1/3
(B) 1/2
(C) 5/8
(D) 2/3
(E) 7/8

1. The sum of two integers to be even, both of them must be even or both of them must be odd. Therefore, the probability of the sum to be even is:

P(the even sum) = P(even, even) + P(odd, odd) = 1/2*1/2 + 1/2*1/2 = 1/2.

2. The product of two integers to be even, either of them or both must be even. Therefore, the probability of the product to be even is:

P(the even product) = P(even, odd) + P(odd, even) + P(even, even) = 1/2*1/2 + 1/2*1/2 + 1/2*1/2 = 3/4.

In half of the cases we have summation and in half of the cases we have multiplication, therefore the overall probability is:

P(even result) = 1/2*1/2 + 1/2*3/4 = 5/8.

Answer: C.

Or another approach.

Let's calculate the probability that the result is odd and subtract that from 1.

1. The sum of two integers to be odd, one of the integers must be odd and another even. Therefore, the probability of the sum to be odd is:

P(the odd sum) = P(even, odd) + P(odd, even) = 1/2*1/2 + 1/2*1/2 = 1/2.

2. The product of two integers to be odd, both must be odd. Therefore, the probability of the product to be odd is:

P(the odd product) = P(odd, odd) = 1/2*1/2 = 1/4.

In half of the cases we have summation and in half of the cases we have multiplication, therefore the overall probability that the result is odd is:

P(odd result) = 1/2*1/2 + 1/2*1/4 = 3/8.

Therefore, the probability that the result is even is:

P(even result) = 1 - P(odd result) = 1 - 3/8 = 5/8.

Answer: C.

Hope it's clear.

Dhruv120292 · Answer

I just took all the 8 cases which are:
case 1- odd odd (*)
Case 2 - Odd odd (+)
Case 3 - Even Even (+)
Case4 - Even Even (*)
Case 5 - Odd Even (+)
Case 6 Odd Even (*)
Case 7 Even odd (+)
Case 8 Even Odd (*)

Therefore total outcomes are 8 and out of this you get even in 5 cases. So Prob should be 5/8. 
Can someone validate this?

bumpbot · Answer

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

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