Hard Manhattan Problem

its easier than you think.!
C) 5/8

total cases when two nos. are randomly picked:
if both are>> odd,odd=sum> EVEN
product> ODD
odd,even=sum> ODD
product> EVEN
even,even=sum> EVEN
product> EVEN
even,odd=sum> ODD
product> EVEN
hence favorable cases= 5
total cases= 8
probability = 5/8

all the best.!

That is not correct OA is C

kakz 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

P(even) = 1 - P(odd).

The following are ways to get an ODD result.

Case 1: odd*odd = odd.
P(odd number and multiplication and odd number) = 1/2 * 1/2 * 1/2 = 1/8.

Case 2: odd+even = odd OR even+odd = odd.
P(odd number and addition and even number) = 1/2 * 1/2 * 1/2 = 1/8.
P(even number and addition and odd number) = 1/2 * 1/2 * 1/2 = 1/8.
1/8 + 1/8 = 2/8.

P(even) = 1 - (1/8 + 2/8) = 5/8.

The correct answer is C.

GMATGuruNY wrote:

kakz 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

P(even) = 1 - P(odd).

The following are ways to get an ODD result.

Case 1: odd*odd = odd.
P(odd number and multiplication and odd number) = 1/2 * 1/2 * 1/2 = 1/8.

Case 2: odd+even = odd OR even+odd = odd.
P(odd number and addition and even number) = 1/2 * 1/2 * 1/2 = 1/8.
P(even number and addition and odd number) = 1/2 * 1/2 * 1/2 = 1/8.
1/8 + 1/8 = 2/8.

P(even) = 1 - (1/8 + 2/8) = 5/8.

The correct answer is C.

Hi forgive my stupidity. Could you further explain the part where: "P(odd number and multiplication and odd number) = 1/2 * 1/2 * 1/2 = 1/8" where did the third 1/2 come from?

Thank you!

Mr.Hollywood wrote:

GMATGuruNY wrote:

kakz 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

P(even) = 1 - P(odd).

The following are ways to get an ODD result.

Case 1: odd*odd = odd.
P(odd number and multiplication and odd number) = 1/2 * 1/2 * 1/2 = 1/8.

Case 2: odd+even = odd OR even+odd = odd.
P(odd number and addition and even number) = 1/2 * 1/2 * 1/2 = 1/8.
P(even number and addition and odd number) = 1/2 * 1/2 * 1/2 = 1/8.
1/8 + 1/8 = 2/8.

P(even) = 1 - (1/8 + 2/8) = 5/8.

The correct answer is C.

Hi forgive my stupidity. Could you further explain the part where: "P(odd number and multiplication and odd number) = 1/2 * 1/2 * 1/2 = 1/8" where did the third 1/2 come from?

Thank you!

Case 1: odd*odd = odd.
P(first number is odd) = 1/2. (Since 1/2 of the integers between 1 and 100, inclusive, are odd.)
P(multiplication) = 1/2. (Since the integers are either multiplied or added, with an equal chance of either operation.)
P(second number is odd) = 1/2.
Since we want all of these events to happen together, we multiply the fractions:
1/2 * 1/2 * 1/2 = 1/8.

The same reasoning was applied to the other cases.

1. With multiplication, we get an even result whenever at least one of the numbers is even.

Let A denote the first number being even
Let B denote the second being even

P(A or B) = P(A) + P(B) - P(A and B) = 1/2 + 1/2 - 1/4 = 3/4


2. With addition, we get an even result whenever both numbers are even OR both numbers are odd.

Keeping with the same notation above...

P(A and B) = 1/4
P(not A and not B) = 1/4

Since the above possibilities are mutually exclusive, we simply add them together to get the probability of one or the other occurring.

Finally, we average the two cases together, since multiplication or addition can occur with equal chance: (3/4 + 1/2) / 2 = 5/8.

Mr.Hollywood wrote:

GMATGuruNY wrote:

kakz 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

P(even) = 1 - P(odd).

The following are ways to get an ODD result.

Case 1: odd*odd = odd.
P(odd number and multiplication and odd number) = 1/2 * 1/2 * 1/2 = 1/8.

Case 2: odd+even = odd OR even+odd = odd.
P(odd number and addition and even number) = 1/2 * 1/2 * 1/2 = 1/8.
P(even number and addition and odd number) = 1/2 * 1/2 * 1/2 = 1/8.
1/8 + 1/8 = 2/8.

P(even) = 1 - (1/8 + 2/8) = 5/8.

The correct answer is C.

Hi forgive my stupidity. Could you further explain the part where: "P(odd number and multiplication and odd number) = 1/2 * 1/2 * 1/2 = 1/8" where did the third 1/2 come from?

Thank you!

I came across this question today and just stumbled upon this great explanation. I just have one quick question. How can we say that the probability of selecting the second odd number is also 1/2? Assuming we have 50 odd numbers in a group of 100 integers, shouldn't the probability of randomly selecting the second odd number be 49/99? Instead of 1/2?

Thanks!

Kyan wrote:

Mr.Hollywood wrote:

GMATGuruNY wrote:

kakz 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

P(even) = 1 - P(odd).

The following are ways to get an ODD result.

Case 1: odd*odd = odd.
P(odd number and multiplication and odd number) = 1/2 * 1/2 * 1/2 = 1/8.

Case 2: odd+even = odd OR even+odd = odd.
P(odd number and addition and even number) = 1/2 * 1/2 * 1/2 = 1/8.
P(even number and addition and odd number) = 1/2 * 1/2 * 1/2 = 1/8.
1/8 + 1/8 = 2/8.

P(even) = 1 - (1/8 + 2/8) = 5/8.

The correct answer is C.

Hi forgive my stupidity. Could you further explain the part where: "P(odd number and multiplication and odd number) = 1/2 * 1/2 * 1/2 = 1/8" where did the third 1/2 come from?

Thank you!

I came across this question today and just stumbled upon this great explanation. I just have one quick question. How can we say that the probability of selecting the second odd number is also 1/2? Assuming we have 50 odd numbers in a group of 100 integers, shouldn't the probability of randomly selecting the second odd number be 49/99? Instead of 1/2?

Thanks!

The two integers are INDEPENDENTLY chosen.
This means that neither selection restricts the other.
The first number selected can be any integer between 1 and 100, inclusive, as can the second number.
Thus -- in each case -- the probability of selecting an odd integer = 1/2.