A set has ten positive integers six even and four odd. Two integers

Question

A set has ten positive integers six even and four odd. Two integers are picked randomly and added. What is the probability that the sum is even?

A.  \frac{1}{2}

B.  \frac{7}{15}

C.  \frac{8}{15}

d.  \frac{2}{3}

E.  \frac{3}{5}

gmatophobia · Answer

For the sum to be even, either both the terms chosen must be even or both the terms chosen must be odd.

Probability that both the terms are even =  \frac{^6C_2 }{ ^{10}C_2}

Probability that both the terms are odd=  \frac{^4C_2 }{ ^{10}C_2}

Net Proabiblity =  \frac{^6C_2 }{ ^{10}C_2}  +  \frac{^4C_2 }{ ^{10}C_2}

Solving we get  \frac{7}{15}

Option B

Iserhard · Answer

Hello, one question

Can we do 1 - Prob (EO)?

Idk what Im doing wrong here.

1 - (4/10 * 6/9) =

(1 - 4/15) = 3/5

Cheers

ManifestDreamMBA · Answer

For sum to to be even, either both numbers need to be even or both need to be odd.

Case 1: Both Even = 6C2 = 15
Case 1: Both Odd = 4C2 = 6
Total Cases = 10C2 = 45
Favourable Cases = 15+6 = 21
P = 21/45 = 7/15

Or doing it your way, we pick 1 even and 1 odd = 6C1 * 4C1 = 24
Total Cases = 10C2 = 45
P(Sum odd) = 24/45 = 8/15
P(Sum Even) = 1-8/15 = 7/15

tx5008 · Answer

Okay. I have another way for this one.

Using simple/random numbers, I quickly calculated to the side, what kind of numbers (even and odd) added together will give me an even sum. I determined (1) even # + (1) even # = even number and (1) odd # + (1) odd # = even number.

1. Set up slots - (1) even # + (1) even # = even number CASE:

6 (chances i pick an even number)/10 (total number of integers in the set) * 5 (chances i pick an even number NOW - since I am not replacing my first pick)/9 (total number of integers NOW in the set) = 15/45

2. Set up slots - (1) odd # + (1) odd # = even number CASE:

4 (chances i pick an odd number)/10 (total number of integers in the set) * 3 (chances i pick an odd number NOW - since I am not replacing my first pick)/9 (total number of integers NOW in the set) = 6/45

3. Total probability (Addition)

15/45 + 6/45 = 21/45 =  7/15!!

Tattvamasi · Answer

Bunuel  the stem does not specify whether selection is with or without replacement (i.e., whether repetition is allowed).
How should we interpret this? Is it standard to assume that repetition is not allowed (selection without replacement) unless otherwise stated? Is that the correct approach to arrive at the answer for such questions?

Bunuel · Answer

Proper GMAT questions always make it clear whether selection is with or without replacement when that detail matters. It would’ve been better if that were clarified here too. However, in this case, the numbers are picked together to be added, so repetition wouldn’t occur, it’s reasonable to assume selection without replacement.

A set has ten positive integers six even and four odd. Two integers

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A set has ten positive integers six even and four odd. Two integers

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