Odd Even Probabilities

There are 5 odd and 5 even numbers between 1 and 10.

We can get an even sum, iff we add either two even or two odd numbers.

We can get an odd sum, iff we add an odd and an even number (in any order).

If we pick an even number first, then we can choose the second (distinct) number out of 4 even and 5 odd numbers (and vice versa).

Therefore, the probability of getting an odd sum:

$p( S_\text{odd} ) = p(even, odd) + p(odd, even) = \frac {5}{10} × \frac {5}{9} + \frac {5}{10} × \frac {5}{9} = \frac {5}{9}$

And the probability of getting an even sum:

$p( S_\text{even} ) = 1 - p( S_\text{odd} ) = 1 - \frac {5}{9} = \frac {4}{9}$

Hence, it is more likely to get an odd sum and our answer should be:

$\boxed { \text {No} }$

There are $\binom {10} 2 = 45$ ways to choose two cards, with equal probability.

There are $\binom 5 2 = 10$ ways in which they are both even, and $\binom 5 2 = 10$ ways in which they are both odd; in total, this gives $10 + 10 = 20$ ways in which the sum is even.

On the other hand, there are $\binom 5 1 \cdot \binom 5 1 = 5 \cdot 5 = 25$ ways to choose one odd and one even card, with an odd sum.

Since $25 > 20$ , the probability of an odd sum is greater.

When you can pick a card twice, there are equal cases of odd and even sums (because of symmetry). To get the set of the valid cases we should remove all the same card cases from this set. The removed cases have even sum, so there are more odd cases remaining.

Even + Even = Even

Odd + Odd = Even

Even + Odd = Odd

Odd + Even = Odd

It is more likely to get an even and an odd card than it is to get cards with the same parity (even and even, or odd and odd). Therefore, it is more likely to get an odd sum.

If you toss out an even card, there are more odd cards in the pile, so it is more likely to get another odd card, which means it is more likely to get an odd sum

For a pack of $6$ cards, the probability to get cards with different parities is $3/5$ (After you choose a card, 3 of the 5 cards will be of a different parity to the card you chose). The probability to get cards with same parities is $2/5$ because you already chose a card of the same parity, therefore it is less likely to get another one.

As the problem states, the probability of getting an even or odd card are equal at first. However, when choosing the second card, the first card is not available, so a card of the opposite parity is more likely to get chosen the second time.

If both cards have the same parity (less likely), the sum will be even. If they have different parity (more likely), the sum will be odd. Therefore an odd sum is more likely than an even sum, and the answer is $\boxed{\text{no}}$ .

To get an even sum, the two cards must have the same... let's call it "parity". By this word I mean that the cards must either both be even or both be odd. Now, there are 5 cards of each parity. However, for each card from A-10, if you want to seek a "mate" with the same parity.... well, a card can't mate with itself, and since any card has the same parity has itself, you suddenly only have 4 cards with the same parity to choose from! So that's 10 x 4 = 40 pairs with the same parity, and thus 40 possible pairs with even sum.

On the other hand, to get an odd sum the cards the cards must have opposite parity instead. For each of the 10 values, the number of prospective partners with opposite parity is 5, since -- by virtue of no card having an opposite parity of itself -- it can choose from all the cards with that parity in the original set, which is half of 10, which is 5. So there are 5x10 = 50 pairs with opposite parity, and that in turn gives us 50 possible pairs with an odd sum.

In conclusion: there is only a chance of 40/90 or 4/9 to get a pair with an even sum , while there is a 50/90 or 5/9 probability of getting two cards with an odd sum . Thus, since the chance of getting an odd sum are higher, the answer is a resounding "no".

Let's call the Even Numbers E, and the Odd Numbers O. If we're summing 2 of these cards, the 3 cases that can happen are as follows:

Because we have the same number of odds and evens, looking at our outcomes above, we can see that the probability is not equal, but favors the Equal Sum.

There are 45 ways of drawing two cards randomly. Because 45 is an odd number, there cannot possibly be equal probablities of even and odd sum.

If we pick 2 cards, we have 3 possibilities:
1) 2 odd numbers, the sum of 2 odd numbers is another odd.
2) 2 even numbers, the sum of 2 even numbers is a odd number.
3) 1 odd and 1 even, where the sum is a even number
So we will have 2/3 of possibilities of take a odd number from the sum of the 2 cards.

If we add two even or two odd numbers in both case the sum is even.And if we add an odd and an even number the sum is an odd number hence the probability of even and odd sum is 2/3 and 1/3 respectively

Its simple when you know the following rules:

1) Odd + Odd = Even.

2) Even + Even = Even.

3) Odd + Even = Odd.

The sum of two positive integers is either even or odd.

Since $\begin{pmatrix} 10 \\ 2 \end{pmatrix} = 45$ and $45 \neq 0\pmod{2}$ there are not the same number of even sums as there are odd sums.

Probability of picking up a even number 1st is 1/5 and then another even number is 1/4, but if in second case we draw an odd number probability is 1/5.Hence the probability differs.

So if we get 2 even numbers then the sum will be an even number if we get 2 odd numbers then the sum will be an even number but if we get 1 odd and 1 even then the sum will be an odd number hence there is a higher chance of the sum to be an even number than an odd number

On the first draw, it's 50/50 chance of getting an odd number vs an even number. On the 2nd draw (without replacement), will be more likely to be odd if your first draw was as even, and vice versa. An odd plus and even, an even plus an odd. Those are the most likely outcomes.

Odd number + odd number = even number. Even number + even number = even number. Odd number + even number = odd number. Hence the probability of getting an odd number is 1/3 while the probability of getting an even number is 2/3. Hence, they are not equal.

If I first draw an even card, there is a 5/9 probability that the second card is odd, yielding an odd sum. If I first draw an odd card, there is a 5/9 probability that the second card is even, yielding an odd sum. Thus the probability of an odd sum is 5/9.

The sum of two evens is always even, the sum of two odds is always odd, the sum of an even and an odd is always odd, similar to multiplying positives and negatives.

If you choose an even number on the first go, there will be less evens for you to choose on the second go, and vis versa with odds, therefore you are more likely to choose two numbers with different parity, meaning the sum of the two numbers is more likely to be odd.

There are 100 possible outcomes of picking two numbers between 1-10 (with replacement).

50 have an odd sum, 50 have an even sum.

However, without replacement 10 outcomes, the diagonal of (n,n) combinations, all even sums, become impossible outcomes.

This reduces the sample space to 90 outcomes, 50 odd, 40 even.

The probability of odd outcome is then 5/9 against 4/9 for even.

The result follows from that ${10\choose2} =45$ is odd.

Without making a complicated formula, randomly choosing two distinct cards eliminates the possibility of doubles to make some even number sums ( can’t choose 5 twice to make 10) so this means there are more odd number combinations.

For odd sum one card to be odd chosen from the 5 available. And the other card to be even chosen out of 5 available even cards... = 5c1 * 5c1 = 25; Probability = 25/ (10c2) =25/45 = 5/9

For even sum, either two cards to be chosen from 5 available odd cards = 5c2 = 10. Or they are to be chosen from even cards.. 10+10=20; Probability = 20/45 =4/9

*Note that these are mutually exclusive events... thus their probabilities add up to 1.