First 15 positive integers are written on a board. If two numbers are

Question

First 15 positive integers are written on a board. If two numbers are selected one by one from the board at random (the numbers are not necessarily different), what is the probability that the sum of these numbers is odd?

A) 49/225
B) 56/225
C) 98/225
D) 105/225
E) 112/225

A) 49/225
B) 56/225
C) 98/225
D) 105/225
E) 112/225

dabaobao · Answer

Veritas Prep Official Solution

Numbers: 1, 2, 3, 4, …, 13, 14, 15

When will the sum of two of these numbers be odd? When one number is odd and the other is even.

P(Sum is Odd) = P(First number is Odd)*P(Second number is even) + P(First number is Even)*P(Second number is Odd)

P(Sum is Odd) = (8/15)*(7/15) + (7/15)*(8/15) = 112/225

Explanation

On first selection, we can pick any number so the probability is 1. The second selection depends on what you selected in the first pick. If you selected an odd number in the first pick, the probability of selecting an even number is 7/15. If you selected an even number in the first pick, the probability of selecting an odd number is 8/15. So what do you do? Do you use 7/15 or 8/15 with 1? You cannot say so you must take individual cases.

Case 1: Select an odd number and then an even number: (8/15) * (7/15)

Case 2: Select an even number and then an odd number: (7/15) * (8/15)

The total cases considered here are 15*15 (select first number in 15 ways and select the second number in 15 ways since the second number can be the same as the first number). In 8*7 ways, you will select an odd number and then an even number. In 7*8 ways, you will select an even number and then an odd number. In both the cases, the sum will be odd. This gives us a probability of (56+56)/225 = 112/225

The total probability of 1 is obtained as follows:

1 = P(first number odd, second number even) + P(first number even, second number odd) + P(first number odd, second number odd) + P(first number even, second number even)

= 56/225 + 56/225 + 64/225 + 49/225 = 1

We are only interested in the 56/225 + 56/225 part.

itspC · Answer

Why so complex solution ? What I did:

1-(P(odd+odd)+ P(even+even)=
1-49/225-64/225 = 112/225

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vaibhav1221 · Answer

another easy way to solve for this is in by looking at the problem in the following way

odd + odd = Even
even + even = Even
odd + even = Odd
even + odd = Odd.

This implies that there is a 50% probability of a number being odd when any two numbers are added.

This would help approximate your answer to 112/225.

Had there been an option with 113/225 as the probable answer, this method would have rendered useless.

gvvsnraju@12 · Answer

VeritasKarishma  :

i have a query in the question.

how do we know that

1. "one by one" means with replacement (or) without replacement.
2. What is the significance of "the numbers are not necessarily different"

Thanks in advance
Raju

KarishmaB · Answer

One by one only implies that there is a first pick and a second pick. It would be specifically mentioned if the picks were with replacement. We would be given "a number is picked and put back and then another number is picked". Hence, if it is not given that the selection is put back, we assume that it is not put back.
Here, the question tells us that the same number can be picked again so this is a way of saying that it is with replacement. So you don't pick away the number from the list - you just select one and write down. Then from the same list of 15 you select another and write down (so same number can be picked again)
Whether it is with replacement or without replacement, you will know from context of the question.

ScottTargetTestPrep · Answer

Of the first 15 positive integers, 8 are odd and 7 are even. The only way to not get an odd sum would be if we select 2 odd numbers or 2 even numbers.

P(2 even numbers) = 7/15 x 7/15 = 49/225.

P(2 odd numbers) = 8/15 x 8/15 = 64/225

Therefore, P(odd sum) = 225/225 - 49/225 - 64/225 = 112/225.

Alternate solution:

Of the first 15 positive integers, 8 are odd and even. The only way to get an odd sum would be if the first number is odd and the second is even or the first number is even and the second is odd.

P(1st = odd, 2nd = even) = 8/15 x 7/15 = 56/225.

P(1st = even, 2nd = odd) = 7/15 x 8/15 = 56/225

Therefore, P(odd sum) = 56/225 + 64/225 = 112/225.

Answer: E

Archit3110 · Answer

total odd integers 1 to 15 ; 8
and total even integers 2 to 14 ; 7
total ways to get sum of 2 integers as odd ; odd + even
8/15 * 7/15 * 2c1 = 112/225
option E

dave13 · Answer

KarishmaB in the second (alternate) solution above why do we consider 2 cases (even; odd) and (odd ;even) isnt the final result the same where order doesnt matter ? in which case i wouldnt need to consider those arrangements , using probability method. btw does probability method arrange or unarrange ? thanks!

for example in this question https://gmatclub.com/forum/from-the-5-p ... 79784.html you dont consider cases i.e. you dont multiply 1/10 by 3! can you explain paradox ?

KarishmaB · Answer

Probability gives you the probability of each defined selection.

I have 15 numbers. Out of them, I pick an odd number with the probability 8/15. Next, I pick another number and that is even. Probability of picking even is 7/15.
Overall probability of picking odd and then even in two picks is 8/15 * 7/15 = 56/225.
But let's not forget the case in which we pick even first and then odd. That will be a different case though it will give the same overall probability. 
Or we can say that the probability of picking odd and then even is 56/225 but we can arrange it as first even and then odd too so let's multiply it by 2.

This is exactly what we do when we need say exactly 2 heads on 3 tosses of a fair coin.

We say the probability of Heads, Heads, Tails (in that order) is 1/2 * 1/2 * 1/2 = 1/8. 
But these do not include cases Heads, Tails, Heads etc so we need to arrange HHT which we can do in 3!/2! = 3 ways. 
So we multiply 1/8 by 3 to "arrange" HHT.

dave13 · Answer

KarishmaB thanks! following the same logic in the question below why then we dont multiply 1/10 by 3! https://gmatclub.com/forum/from-the-5-p ... 79784.html

chiplesschap · Answer

Can someone explain what's wrong with using combinations here.

I did (8C1 * 7C1)/(15C2) + (7C1 * 8C1)/(15C2)?

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Bunuel · Answer

15C2 gives the number of pairs of  different  numbers from 1, 2, 3, ..., 15. However, the numbers can be the same, so the denominators should be 15*15 instead.

chiplesschap · Answer

So if the stem didn't explicitly say "the numbers are not necessarily different" and if I use 15C2 for denominator, the probability would exceed 1. What am I missing? Thanks for the response!

Bunuel · Answer

If the numbers could not be the same, then you can do either 8/15*7/14*2 = 8/15 or 8C1*7C1/15C2 = 8/15 (here, you don't multiply by 2 because the formula 8C1*7C1 = 56 gives all unordered combinations of one odd and one even number, and the denominator, 15C2, also gives the number of unordered pairs, so these two balance each other out).

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