Question

Five cards are drawn randomly from a standard deck of 52 cards.

Answer 1

Answer:  0.00174

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Explanation:

There are 4 ways to select 3 aces where order doesn't matter. It's basically the same as saying there are 4 ways to leave out one ace.

Then we have 2 slots left to fill. There are (48*47)/2 = 1128 ways to do this where order doesn't matter. The 48 is from the fact there are 52-4 = 48 cards that aren't aces. We step down to 47 after picking the first non-ace card. The 2 in the denominator is to correct for double counting.

We found there were 4 ways to pick the three aces, and 1128 ways to pick the other two non-ace cards. Overall, there are 4*1128 = 4512 ways to pick all five cards where we have exactly 3 aces.

This is out of 52C5 = 2,598,960 ways to select any five cards from a 52 card deck. I'm using the nCr formula which is

$_n C _r = \frac{n!}{r!*(n-r)!}$

Use n = 52 and r = 5 to get the value mentioned. The exclamation marks indicate factorial.

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To recap, there are

• 4512 ways to pick exactly three aces
• 2,598,960 ways to pick five cards without any restrictions

Dividing the two values gets us the final answer

4512/(2,598,960) = 0.001736079047

This value rounds to 0.00174

Answer 2

Answer:

0.00694

Step-by-step explanation:

Probability ( First 3 cards are aces and the next 2 are not aces)

= 4/52 * 3/51 * 2/50 * 47/49 * 46/48

= 1/13 * 1/17 * 1/ 25 * 47/49 * 46/48

= 2256/12994800

= 0.00016637424

There are 4 aces in a pack of cards so there are 4 ways of picking these 3 aces

- and there are  5C3 ways of arranging  3 aces in the 5 cards drawn

= 10 ways.

So the required probability =  4 * 10 * 0.00016637424

= 0.00667