I will list two ways of solving this problem.

Let's look at all possible ways of getting two face cards, one even card and one odd card.

First we have to pick the two locations that the two face cards can occupy. This we can do in _4 C_2 = 6 ways. Then the even card and odd card can occupy the other two places in two ways. Therefore, the total number of ways of getting two face cards, one even card and one odd card is 12 .

The probability of drawing a face card is \frac{2}{13} , an even card is \frac{5}{13} and an odd card is \frac{5}{13} .

Therefore, total probability is 12 \times \left( \frac{2}{13} \right)^2 \left( \frac{5}{13} \right)^2 .

The other way to do is:

The total number of possible outcomes is given by 52 \times 52 \times 52 \times 52 = 52^4 .

The total number of favorable outcomes is given by 12 \times 12^2 \times 20 \times 20 .

Therefore, probability = \frac{\text{total number of favorable outcomes}}{\text{total number of possible outcomes}} = \frac{12 \times 12^2 \times 20^2}{52^4} = 12 \times \left( \frac{2}{13} \right)^2 \left( \frac{5}{13} \right)^2 .

I found an answer from www.bbc.com

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Probabilities **can be** written as fractions, decimals or percentages on a scale from
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What is the **probability** of selecting a vowel at **random** from the word
**PROBABILITY**?

For more information, see **Probability** - Edexcel - Revision 2 - GCSE Maths - BBC Bitesize

I found an answer from www.khanacademy.com

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I found an answer from www.britannica.com

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I found an answer from www.nationalgeographic.com

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I found an answer from www.quora.com

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I found an answer from math.stackexchange.com

Finding **probability of 2 cards** summing to **an odd** number ...

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I found an answer from en.wikipedia.org

Phase 10 - Wikipedia

**Random chance**, High. Skill(s) required, Saving important **cards**; knowing when
to put down those **cards**; matching, ordering. Phase 10 is a **card** game created in
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I found an answer from www.scientificamerican.com

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