A probability problem by Nicholas Tomlin

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 $\frac{1}{\binom{10}{5}} = \frac{1}{252}$ , and therefore, the probability that the numbers are even is equal to $1 - \frac{1}{252} = \frac{251}{252}$ , and $a + b = 251 + 252 = \boxed{503}$ .

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

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

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 $\frac{a}{b} = \frac{251}{252}$ 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. $|A'| = |U| - |A|$ Where $|A|$ is the number of ways that the product is even. Then $|A'|$ 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.

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

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

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 $^{10}C_5=252$ ways and $^{10}C_5 -1=251$ of them gives even product. So $\frac{a}{b}=\frac{251}{252}$

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

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

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

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

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) )

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 $\large (1,3,5,7,9)$ , there is only one way to make a set of 5 integers such that the product is odd. Hence our probability is:

$\large P(Even)=1-P(Odd)=1-\frac{1}{{10 \choose 5}}=\frac{251}{252}$

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

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

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

$even \times even = even$

$even \times odd = even$

$odd \times odd = odd$

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

$251 + 252 = \boxed{503}$

Since product is invariant under the order of the 5 chooses, this reduces down to the problem of counting the number of combinations of $S$ 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 $\{1,3,5,7,9\}$ , therefore the likelihood is $1 - \frac{1}{10\choose 5}$ , and since $b-1$ and $b$ are guaranteed coprime, the answer is just $a + b = 2{10\choose 5}-1 = \boxed{503}$ .

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 $5$ -number sets chosen is equal to ${10 \choose 5}$ = $252$ . But hey! Don't start doing so many combinations. The only way that the product of $5$ numbers from that set is odd, is if we choose $5$ odd numbers! So we have only $5$ 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 $252 - 1 = 251$ . The probability is $\frac {251}{252}$ . Thus, $a + b = 251 + 252 = \boxed {503}$ .

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.

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:

$\frac{(10!)}{(5!)(5!)} = 252$ 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:

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

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