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A card is drawn at random without replacement. A second card is then drawn. Determine the probability of getting a number greater than 9 and a number less than 4 if cards are selected at a random​

User SvenK
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1 Answer

6 votes

Explanation:

it does not say what the card deck consists of, and how many cards of what type there are.

so, I assume a standard playing card set of 52 cards (no jokers) with 4 suits of

Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K

in that ranking of value.

so there are 4 cards of each of the 13 different types.

a card with a number greater than 9 means a 10, a J, a Q, or a K.

so, in total there are 4×4 = 16 such cards in the deck.

a card with a number less than 4 means an Ace, a 2, or a 3.

so, in total there are 4×3 = 12 such cards in the deck.

to get a number greater than 9 and a number less than 4 is a combination of the first card being greater than 9 and the 2 card being less than 4 OR the first card being less than 4 and the second card being greater than 9.

an "and" operation (without overlap) we represent by a multiplication, and an "or" operation (without overlap) by an addition.

remember, a probability is always

"desired cases / total possible cases".

so, the probability to draw a first card greater than 9 is

16/52 = 4/13

the probability to draw a second card greater than 9 (after a first card less than 4) is

16/51

because we still have all the large cards, but one of the smaller ones is now gone.

the probability to draw a first card less than 4 is

12/52 = 3/13

the probability to draw a second card less than 4 (after a first card greater than 9) is

12/51.

so, the complete probability of the scenario is

4/13 × 12/51 + 3/13 × 16/51 = 48/663 + 48/663 = 96/663 =

= 32/221 = 0.14479638...

if my assumptions about the cards are wrong, please adapt the numbers accordingly. but the basic structure of the calculation is the same.

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