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Number N is randomly selected from a set of all primes betwe [#permalink]

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01 Aug 2014, 20:21

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Number N is randomly selected from a set of all primes between 10 and 40, inclusive. Number K is selected from a set of all multiples of 5 between 10 and 40 inclusive. What is the probability that N+K is odd?

Re: Number N is randomly selected from a set of all primes betwe [#permalink]

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01 Aug 2014, 20:53

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To get an odd number from two values being added together, one value must be even and the second value must be odd. For number \(N\) for primes between 10 and 40, there are 7 values (13, 17, 19, 23, 29, 31, 37) which are all odd. However, this is not even necessary since all you need to know is that 2 is the even prime and all primes between 10 and 40 are odd. Therefore, the only way \(K+M\) can be odd is if the number \(K\) is even. There is a total of 7 values of \(K\), with 4 even values and 3 odd values.

Therefore, the probability is \(\frac{4}{7}\)which is D

Number N is randomly selected from a set of all primes between 10 and 40, inclusive. Number K is selected from a set of all multiples of 5 between 10 and 40 inclusive. What is the probability that N+K is odd?

(A) 1/2 (B) 2/3 (C) 3/4 (D) 4/7 (E) 5/8

All primes except 2 are odd, thus N must be odd. For N + K = odd + K to be odd, K must be even.

There are 7 multiples of 5 between 10 and 40, inclusive: 10, 15, 20, 25, 30, 35, and 40. Out of them 3 (15, 25, and 35) are odd.

Therefore the probability that N + K is odd is 4/7.

Re: Number N is randomly selected from a set of all primes betwe [#permalink]

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25 May 2015, 11:46

Bunuel wrote:

kennaval wrote:

Number N is randomly selected from a set of all primes between 10 and 40, inclusive. Number K is selected from a set of all multiples of 5 between 10 and 40 inclusive. What is the probability that N+K is odd?

(A) 1/2 (B) 2/3 (C) 3/4 (D) 4/7 (E) 5/8

All primes except 2 are odd, thus N must be odd. For N + K = odd + K to be odd, K must be even.

There are 7 multiples of 5 between 10 and 40, inclusive: 10, 15, 20, 25, 30, 35, and 40. Out of them 3 (15, 25, and 35) are odd.

Therefore the probability that N + K is odd is 4/7.

Between 10 and 40 there are 8 Prime Numbers, and between 10 and 40 there are 7 multiples of 5, of which 3 are odd and 4 are even. We need N + K = odd, so we have 4 in our favor out of 7 multiples. but my question is, why not any number from Prime Numbers set is participating in total and favorable number of cases?

Something like - Total Number of cases = 8 (primes) + 7(all multiples of 5) = 15 Favorable = 8 (All Primes, because they all are odd and will result in Odd when added to even) + 4 (Even Multiples of 5) = 12 Prob = 12/15 = 3/4 (Why not this answer?)

I am thinking about this, because question is saying N is randomly selected from a set of all primes....?

Number N is randomly selected from a set of all primes between 10 and 40, inclusive. Number K is selected from a set of all multiples of 5 between 10 and 40 inclusive. What is the probability that N+K is odd?

(A) 1/2 (B) 2/3 (C) 3/4 (D) 4/7 (E) 5/8

All primes except 2 are odd, thus N must be odd. For N + K = odd + K to be odd, K must be even.

There are 7 multiples of 5 between 10 and 40, inclusive: 10, 15, 20, 25, 30, 35, and 40. Out of them 3 (15, 25, and 35) are odd.

Therefore the probability that N + K is odd is 4/7.

Between 10 and 40 there are 8 Prime Numbers, and between 10 and 40 there are 7 multiples of 5, of which 3 are odd and 4 are even. We need N + K = odd, so we have 4 in our favor out of 7 multiples. but my question is, why not any number from Prime Numbers set is participating in total and favorable number of cases?

Something like - Total Number of cases = 8 (primes) + 7(all multiples of 5) = 15 Favorable = 8 (All Primes, because they all are odd and will result in Odd when added to even) + 4 (Even Multiples of 5) = 12 Prob = 12/15 = 3/4 (Why not this answer?)

I am thinking about this, because question is saying N is randomly selected from a set of all primes....?

The mistake which you have made is a common confusion which students have between OR and AND events.

In case of an OR event we use the addition sign and in case of an AND event we use the multiplication sign. In this question for N + K to be odd you need to select an odd prime number between 10 and 40 inclusive AND an even multiple of 5 between 10 and 40 inclusive. You have rightly found out the number of ways for both the events.

Your process will give the right answer if you consider the events as AND events instead of OR events i.e. use the multiplication sign instead of an addition sign.

So, P(N + K to be odd) = P(N to be odd) AND P(K to be even)

\(= \frac{8}{8} * \frac{4}{7} = \frac{4}{7}\) which is our answer.

Between 10 and 40 there are 8 Prime Numbers, and between 10 and 40 there are 7 multiples of 5, of which 3 are odd and 4 are even. We need N + K = odd, so we have 4 in our favor out of 7 multiples. but my question is, why not any number from Prime Numbers set is participating in total and favorable number of cases?

Something like - Total Number of cases = 8 (primes) + 7(all multiples of 5) = 15 Favorable = 8 (All Primes, because they all are odd and will result in Odd when added to even) + 4 (Even Multiples of 5) = 12 Prob = 12/15 = 3/4 (Why not this answer?)

I am thinking about this, because question is saying N is randomly selected from a set of all primes....?

The mistake which you have made is a common confusion which students have between OR and AND events.

In case of an OR event we use the addition sign and in case of an AND event we use the multiplication sign. In this question for N + K to be odd you need to select an odd prime number between 10 and 40 inclusive AND an even multiple of 5 between 10 and 40 inclusive. You have rightly found out the number of ways for both the events.

Your process will give the right answer if you consider the events as AND events instead of OR events i.e. use the multiplication sign instead of an addition sign.

So, P(N + K to be odd) = P(N to be odd) AND P(K to be even)

\(= \frac{8}{8} * \frac{4}{7} = \frac{4}{7}\) which is our answer.

Hope it's clear

Regards Harsh

Just to add to what Harsh is saying - if one event can occur in m ways and a second can occur independently of the first in n ways, then the two events can occur in mn ways (this is called Principle of Multiplication).
_________________

Re: Number N is randomly selected from a set of all primes betwe [#permalink]

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21 Jun 2015, 12:08

wunsun wrote:

To get an odd number from two values being added together, one value must be even and the second value must be odd. For number \(N\) for primes between 10 and 40, there are 7 values (13, 17, 19, 23, 29, 31, 37) which are all odd. However, this is not even necessary since all you need to know is that 2 is the even prime and all primes between 10 and 40 are odd. Therefore, the only way \(K+M\) can be odd is if the number \(K\) is even. There is a total of 7 values of \(K\), with 4 even values and 3 odd values.

Therefore, the probability is \(\frac{4}{7}\)which is D

YOU MISSED OUT 11 IT IS ALSO A PRIME NUMBER
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Re: Number N is randomly selected from a set of all primes betwe [#permalink]

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19 Nov 2016, 10:06

My approach was:

All primes between 10 and 40 are: 11,13,17,19,23,29,31,37 = 8 numbers. Multiples of 5 between 10 and 40: 10,15,20,25,30,35,40 = 7 multiples, 3 odd and 4 even.

Total number of outcomes = 8*7/2! = 28. (Divided by 2! to eliminate repetition) Total number of possible K + N combinations to give an odd number (We need even+odd, so 4 out of 7) = 8*4/2! = 16.

Re: Number N is randomly selected from a set of all primes betwe [#permalink]

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19 Nov 2016, 19:45

D, 8 primes between 10 n 40, 7 number multiples of 5, between 10 n 40.. For sum of n+k to be odd, k has to be even...that's 10,20,30,40 out of 10,15,20,25,30,35,40

Re: Number N is randomly selected from a set of all primes betwe [#permalink]

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05 Nov 2017, 13:11

kennaval wrote:

Number N is randomly selected from a set of all primes between 10 and 40, inclusive. Number K is selected from a set of all multiples of 5 between 10 and 40 inclusive. What is the probability that N+K is odd?

(A) 1/2 (B) 2/3 (C) 3/4 (D) 4/7 (E) 5/8

All primes except for 2 are odd. Since the prime numbers we're looking for are all greater than 2, they will all be odd. Since the sum of either odd + even or even + odd results in an odd number, and we know that N must all be odd numbers, we know that then we'd be looking for all the even numbers from set K.

The only multiples of 5 that are even are ones that happen to be multiples of 2, as well (basically anything that ends in a 0). Of the numbers from 10 - 40 inclusive that are multiples of 5 (10, 15, 20, 25, 30, 35, 40) only 10, 20, 30, and 40 (4 of the 7 numbers from K) are even. Since only 4 of the 7 possibilities will result in N + K being odd, D is the right answer.

Number N is randomly selected from a set of all primes betwe [#permalink]

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06 Nov 2017, 23:59

kennaval wrote:

Number N is randomly selected from a set of all primes between 10 and 40, inclusive. Number K is selected from a set of all multiples of 5 between 10 and 40 inclusive. What is the probability that N+K is odd?

(A) 1/2 (B) 2/3 (C) 3/4 (D) 4/7 (E) 5/8

The probability of an event happening is equal to

# of Desired Outcomes______ Total # of Outcomes

Here, "Desired Outcomes" are those in which (N + K) = odd number

1) Elements of the probability calculation

N = 11, 13, 17 ,19 ,23, 29, 31, 37 (8 terms)

K = 10, 15, 20, 25, 30, 35, 40 (7 terms)

All N are odd. For (N + K) to be odd, K must be even: (odd + EVEN) = odd

2) Number of POSSIBLE outcomes for N + K?

8 * 7 = 56

3) Number of DESIRED outcomes, where N + K is odd?

All N are odd. 4 of K's terms are even.

So for (N + K) to be odd, the 4 even multiples of five (10, 20, 30, 40), in K, could be paired with all 8 odd primes in N. That means:

4 * 8 = 32 favorable outcomes

4) Probability that N + K is odd? Favorable/Possible:

Number N is randomly selected from a set of all primes between 10 and 40, inclusive. Number K is selected from a set of all multiples of 5 between 10 and 40 inclusive. What is the probability that N+K is odd?

(A) 1/2 (B) 2/3 (C) 3/4 (D) 4/7 (E) 5/8

In order for N + K to be odd, we need an even + odd or odd + even.

Since N is a prime between 10 and 40, N must be odd.

Thus, we need to determine how many even multiples of 5 there are from 10 to 40 inclusive. We have 10, 20, 30, and 40. We also have (40 - 10)/5 + 1 = 7 total multiples of 5 from 10 to 40 inclusive.

Thus, the probability that N is odd and K is even is 1 x 4/7 = 4/7.

Answer: D
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Re: Number N is randomly selected from a set of all primes betwe [#permalink]

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08 Nov 2017, 19:35

kennaval wrote:

Number N is randomly selected from a set of all primes between 10 and 40, inclusive. Number K is selected from a set of all multiples of 5 between 10 and 40 inclusive. What is the probability that N+K is odd?

(A) 1/2 (B) 2/3 (C) 3/4 (D) 4/7 (E) 5/8

Set N = {11,13,17,19,23,29, 31, 37,39} Set K = {10,15,20,25,30,35}

N+K is odd if N is odd and K is even. So, Probability = N+K odd/ N+K total = 9*4 / 9*7 = 4/7

Number N is randomly selected from a set of all primes between 10 and 40, inclusive. Number K is selected from a set of all multiples of 5 between 10 and 40 inclusive. What is the probability that N+K is odd?

(A) 1/2 (B) 2/3 (C) 3/4 (D) 4/7 (E) 5/8

Notice that there are 2 ways that the sum N+K can be ODD 1) N is ODD and K is EVEN 2) N is EVEN and K is ODD

However, N cannot be EVEN, since all of the primes between 10 and 40 are ODD (11, 13, 17, 19, . . . . 31, 37) So, case 2 (above) is impossible

So, P(N+K is odd) = P(N is odd AND K is even) = P(N is odd) x P(K is even)

------ASIDE------- Possible values of N: (11, 13, 17, 19, . . . . 31, 37) So, P(N is odd) = 1

Possible values of K: (10, 15, 20, 25, 30, 35, 40) So, P(K is even) = 4/7 -----------------------------------------

So, P(N+K is odd) = P(N is odd AND K is even) = P(N is odd) x P(K is even) = 1 x 4/7 = 4/7 = D