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

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First 15 positive integers are written on a board. If two numbers are  [#permalink]

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New post 25 May 2019, 12:58
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Question Stats:

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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
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Re: First 15 positive integers are written on a board. If two numbers are  [#permalink]

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New post 25 May 2019, 13:01
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dabaobao wrote:
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



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.
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Re: First 15 positive integers are written on a board. If two numbers are  [#permalink]

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New post 18 Aug 2019, 02:55
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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Re: First 15 positive integers are written on a board. If two numbers are  [#permalink]

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New post 20 Aug 2019, 01:26
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.
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Re: First 15 positive integers are written on a board. If two numbers are  [#permalink]

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New post 21 Nov 2019, 06:02
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
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Re: First 15 positive integers are written on a board. If two numbers are  [#permalink]

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New post 21 Nov 2019, 21:48
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gvvsnraju@12 wrote:
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


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.
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Re: First 15 positive integers are written on a board. If two numbers are  [#permalink]

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New post 24 Nov 2019, 18:58
dabaobao wrote:
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


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
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Re: First 15 positive integers are written on a board. If two numbers are   [#permalink] 24 Nov 2019, 18:58
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