counting/probability with cards

Question

How many ways can you pick an ace followed by a card that is less than the number 5 from a deck of 52 cards
(no replacement)?

Answer 1

This is a no replacement probability problem.

Let's start with the first card draw

P(Ace) = 4 aces / 52 cards = 4/52

 

For the second draw we want card with a number less than 5.

P(card with number less than 5)

= p(any 2 card, any 3 card, any 4 card)

= 3 * 4 / 51

We use 51 since we did not replace the first card drawn.

= 12 / 51

 

Now the combined probability

= p(ace) * p( num < 5)

= 4/52 * 12/51

= 1/13 * 4/17

= 4/ (13*17) = 0.01809

 

Just in case they wanted to count the Ace as number 1 because Ace is sometimes counted as a 1 or 11, then we would have second part to be p( card < 5) =( 3*4 + 3) / 51

=15/51

So the final probability will then be

4/52 * 15/51 = 0.02262

 

Answer 2

P(Ace) = 1/13

 

P(number < 5) ?

First, let's assume that this refers to picking a card bearing the numeral 2 or 3 or 4; in this case, P(number < 5) would be 12 / 51 = 4 / 17.*

 

The resultant (desired) Probability of these two events occurring in succession would be the product (1/13)(4/17) = 0.0181 to 4 Dec. Places.

 

* this is assuming that an Ace is NOTY A 1 because there's the letter A on it, and not a numeral