A probability problem by Nicholas Tomlin

Five distinct numbers are randomly chosen from the set S = { 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 } S = \{0,1,2,3,4,5,6,7,8,9\} and multiplied. The likelihood that the product is even can be written as a b \frac{a}{b} , where a , b a, b are relatively prime positive integers. Find a + b a + b .


The answer is 503.

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17 solutions

Josh Petrin
Dec 24, 2013

The product will be even if any of the numbers selected are even. The probability that all of the numbers chosen are odd is equal to 1 ( 10 5 ) = 1 252 \frac{1}{\binom{10}{5}} = \frac{1}{252} , and therefore, the probability that the numbers are even is equal to 1 1 252 = 251 252 1 - \frac{1}{252} = \frac{251}{252} , and a + b = 251 + 252 = 503 a + b = 251 + 252 = \boxed{503} .

Well, looks like I am the only one who took the longer approach. :P

Nice solution Josh! :)

Pranav Arora - 7 years, 5 months ago
Tom Zhou
Dec 27, 2013

There are a total of ( 10 5 ) \binom{10}{5} to choose 5 numbers when order does not matter. The only case where the product is n o t not even is when all 5 integers are odd, and there is only one possibility of that, namely 1 , 3 , 5 , 7 , 1,3,5,7, and 9 9 in a group. So the probability of an even product is 1 1 ( 10 5 ) = 251 252 = a b 1-\frac{1}{\binom{10}{5}}=\frac{251}{252}=\frac{a}{b} . a + b = 503 a+b=\boxed{503} .

First look at his , odd * odd = odd even * odd = even even * even = even In the list we have 5 odd and 5 even numbers So if we choose any of the even number the product is going to be even. Therefore let us first count the all possible combinations i.e. {10/C/5}= 252 now let us count those cases in which we are not choosing any of the even numbers i.e. {5/C /5} = 1 Therefore in the remaining 251 cases we are going to choose atleast 1 of the even numbers in which the product will be even so the required probability for the product to be even is {251/252} Hence a = 251 and b = 252 Therefore a + b = 503

I forgot 0 is an even number XD nice solution

johnson adeleke - 7 years, 5 months ago

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thank you

Ujjwal Mani Tripathi - 7 years, 5 months ago

0 is even ? =D

Ramin Taghizada - 7 years, 5 months ago

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yes ofcourse 0 is even 0 when divided by any natural number reaturns 0 as remainder so 0is even

Ujjwal Mani Tripathi - 7 years, 5 months ago
Jessica Yung
Dec 27, 2013

The product can only be odd if you choose the numbers 1,3,5,7,9. (1 way)

The number of ways you can choose 5 distinct numbers is 10C5 = 252, so a b = 251 252 \frac{a}{b} = \frac{251}{252} and a+b = 503.

You can find the number of ways n(E) by finding 5 cases, (1E(even)4O(odd), 2E3O, 3E2O, 4E1O, 5E0O, 4any*0), which is absolutely waste of time. XD

Use the complement of an event. A = U A |A'| = |U| - |A| Where A |A| is the number of ways that the product is even. Then A |A'| is the number of ways that the product is odd. (If you're finding a probability, just use P(E') = 1 - P(E)), P(E) is the probability that the product is odd.

  • n ( E ) n(E) = 1 ( 1 , 3 , 5 , 7 , 9 > 1 3 5 7 9 = o d d ) (1,3,5,7,9 -> 1*3*5*7*9 = odd)
  • n ( S ) n(S) = The total of ways to choose 5 no.s from 10 no.s = C ( 10 , 5 ) C(10,5)

Therefore P ( E ) = 1 n ( E ) n ( S ) = 1 1 C ( 10 , 5 ) = 251 252 P(E') = 1 - \frac{n(E)}{n(S)} = 1 - \frac{1}{C(10,5)} = \frac{251}{252} -> Answer = 503 \boxed{503}

Joji Joseph
Dec 27, 2013

Product of numbers will be even if at least one of them is even and product will be odd if all of them are odd.There is only one way(13579) to choose 5 odd numbers so that that the product is odd.Rest of all combinations will produce even product.

We can select 5 numbers in 10 C 5 = 252 ^{10}C_5=252 ways and 10 C 5 1 = 251 ^{10}C_5 -1=251 of them gives even product. So a b = 251 252 \frac{a}{b}=\frac{251}{252}

Therefore a=251,b=252 and a + b = 503 \boxed{a+b=503}

Mohamed Mahmoud
Dec 27, 2013

The number of groups we could get = 10C5 = 252

there is only one group which five an odd product which is {1,3,5,7,9}

so a=251 and b=252--> a+b=503

Oliver Welsh
Dec 27, 2013

Consider the value of P ( odd product ) P(\text{odd product}) . In this case, all of the numbers have to be odd. There are 10 10 numbers, 5 5 of which are odd. The number of ways to pick 5 5 numbers from the set is ( 10 5 ) = 252 {10 \choose 5} = 252 and the number of ways to choose 5 5 numbers which are odd is ( 5 5 ) = 1 {5 \choose 5} = 1 . Hence, P ( odd product ) = 1 252 P(\text{odd product}) = \frac{1}{252} Now we will use the property that P ( A ) = 1 P ( A ) P(A') = 1 - P(A) , hence, \begin{align} P(\text{even product}) &= 1 - P(\text{odd product}) \\ &= 1 - \frac{1}{252} \\ &= \frac{251}{252} \end{align} Hence, we have, a + b = 251 + 252 = 503 a + b = 251 + 252 = \fbox{503}

Venture Hi
Nov 2, 2014

the only odd combo is 1,3,5,7,9 LOL

Mayank Holmes
May 20, 2014

the answer is absolutely wrong ........ the answer includes the cases when the product is zero .......... zero is not an even number ! the correct ans should be 377 ( since the actual probability should be (125/252) )

Muhammad Shariq
Jan 1, 2014

Instead of finding the probability that the product is even, we find the probability that the product is odd. The product is odd iff all of the 5 factors are odd. Since this set only contains 5 odd integers, namely ( 1 , 3 , 5 , 7 , 9 ) \large (1,3,5,7,9) , there is only one way to make a set of 5 integers such that the product is odd. Hence our probability is:

P ( E v e n ) = 1 P ( O d d ) = 1 1 ( 10 5 ) = 251 252 \large P(Even)=1-P(Odd)=1-\frac{1}{{10 \choose 5}}=\frac{251}{252}

Hence a + b = 251 + 252 = 503 \large a+b=251+252=\boxed{503} .

Milly Choochoo
Dec 31, 2013

There are 10 10 items in set S S , and so the number of unordered subsets of 5 5 items/numbers in length can be given by ( 10 5 ) 10 \choose 5 or 10 ! ( 10 5 ) ! 5 ! \frac{10!}{(10-5)!5!} , which is equal to 252 252 .

Now, the rules for multiplying odds and evens are as follows.

e v e n × e v e n = e v e n even \times even = even

e v e n × o d d = e v e n even \times odd = even

o d d × o d d = o d d odd \times odd = odd

The only way then for there to be an odd product out of 5 5 random numbers being multiplied is 1 × 3 × 5 × 7 × 9 1 \times 3 \times 5 \times 7 \times 9 ! Try it out yourself! So if the odds for the product of those 5 5 random numbers being multiplied and being odd are 1 252 \frac{1}{252} , the odds for the product of those 5 5 random numbers being multiplied and being even are 251 252 \frac{251}{252} .

251 + 252 = 503 251 + 252 = \boxed{503}

Lee Gao
Dec 31, 2013

Since product is invariant under the order of the 5 chooses, this reduces down to the problem of counting the number of combinations of S S that has at least one even number in it. Equivalently, we count the 5-combinations that are all odd: in this case there's only { 1 , 3 , 5 , 7 , 9 } \{1,3,5,7,9\} , therefore the likelihood is 1 1 ( 10 5 ) 1 - \frac{1}{10\choose 5} , and since b 1 b-1 and b b are guaranteed coprime, the answer is just a + b = 2 ( 10 5 ) 1 = 503 a + b = 2{10\choose 5}-1 = \boxed{503} .

Ishan Pandey
Dec 29, 2013

There are 5 odd numbers and 5 even numbers . Now, we know that : Even * Odd=Even Thus , the only way we can get an odd multiple is if we choose all 5 odd numbers. This can be done in 1 way. The total number of ways of selecting 5 numbers = 10C2= 252 The total number of favourable cases= 252-1=251 Thus a+b= 251+252=503

The total of possible 5 5 -number sets chosen is equal to ( 10 5 ) {10 \choose 5} = 252 252 . But hey! Don't start doing so many combinations. The only way that the product of 5 5 numbers from that set is odd, is if we choose 5 5 odd numbers! So we have only 5 5 odd numbers, which means that there's only one set that satisfies that condition. Hence, the cases in which we'll get an even product are 252 1 = 251 252 - 1 = 251 . The probability is 251 252 \frac {251}{252} . Thus, a + b = 251 + 252 = 503 a + b = 251 + 252 = \boxed {503} .

We can choose 5 numbers from 10 in 10!/5!5!=252 ways. The only combination where the product is odd is only {1,3,5,7,9}, which is one combination. Therefore, the probability of the product being even is 251/252, the sum being 251+252=503.

Parth Chopra
Dec 27, 2013

This problem can be simplified greatly by first trying to find the likelihood that the product is odd . In order for the product to be odd, all 5 numbers must be odd themselves!

The number of ways we can choose 5 numbers from the set of 10 is equal to:

( 10 ! ) ( 5 ! ) ( 5 ! ) = 252 \frac{(10!)}{(5!)(5!)} = 252 different ways.

Now, back to the original question. How many ways can we pick 5 odd numbers? Well, only 1 way, since there is only 1 way to pick 5 things out of 5 things.

This means that the probability that the product is even is equal to:

1 1 252 = 251 252 1 - \frac{1}{252} = \frac{251}{252}

Therefore, the value of a + b a + b is equal to 251 + 252 251 + 252 or 503 \boxed{503} .

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