Question

Consider a regular deck of 52 playing cards.

(a) Suppose you draw one card randomly from the deck. What is the probability that that card is an ace or a king?

(b) Suppose you draw two cards randomly from the deck. What is the probability that both cards are red?

(c) Suppose you draw four cards randomly from the deck. What is the probability that exactly two cards are aces?

(d) Suppose you draw three cards randomly from the deck. What is the probability that at least one of them is a diamond?

(e) Suppose you draw four cards randomly from the deck. What is the probability that all of them are aces?

Answer 1

Solution:

Part a) Suppose you draw one card randomly from the deck.

What is the probability that a card is an ace or a king?

we have total of 52 playing cards.

We have to draw one card from 52 playing cards, thus number of ways of drawing any one card from 52 playing cards = 52C1 ways

and we have to find the probability that a card is an Ace or a King.

there are 4 Ace cards and 4 King cards

Thus 1 Ace card can be drawn from 4 ace cards in 4C1 ways

similarly, for a King, there are 4C1 ways.

Thus

P( An Ace or a King)

where

Thus

P( An Ace or a King)

P( An Ace or a King)

P( An Ace or a King)

P( An Ace or a King) = 0.1538

Part b) Suppose you draw two cards randomly from the deck. What is the probability that both cards are red?

There are total 26 red cards in 52 playing cards.

P( Both Red) = 26C2 / 52C2

where

Thus

P( Both Red) = 26C2 / 52C2

P( Both Red) = 325 / 1326

P( Both Red) = 0.2451

Part c) Suppose you draw four cards randomly from the deck. What is the probability that exactly two cards are aces?

P( Exactly 2 cards are Aces and 2 are other than Aces) = ....?

P( 2 Aces and 2 other) = ....?

There are 4 ace and 52-4=48 Non-Ace cards.

Thus 2 ace cards can be selected from 4 aces in 4C2 ways and

2 non ace cards can be selected from 48 Non aces in 48C2 ways

and any 4 cards can be selected from 52 cards in 52C4 ways

Thus

P( 2 Aces and 2 other)

where

and

Thus

P( 2 Aces and 2 other)

P( 2 Aces and 2 other)

P( 2 Aces and 2 other)

Part d) Suppose you draw three cards randomly from the deck. What is the probability that at least one of them is a diamond?

P( At least one diamond) = 1 - P( No diamond)

No diamond means all three selected cards are non diamond cards

thus there are 52-13 diamond = 39 Non-Diamond cards.

Thus

P( No Diamond) = P( All 3 cards are Non-Diamond)

P( No Diamond) = 39C3 / 52C3

where

and

Thus

P( No Diamond) = 39C3 / 52C3

P( No Diamond) = 9139 / 22100

P( No Diamond) = 0.4135

Then

P( At least one diamond) = 1 - P( No diamond)

P( At least one diamond) = 1 - 0.4135

P( At least one diamond) =0.5865

Part e) Suppose you draw four cards randomly from the deck. What is the probability that all of them are aces?

P( All four cards are an Ace) = ....?

P( All four cards are an Ace) = 4C4 / 52C4

Thus

P( All four cards are an Ace) = 4C4 / 52C4

P( All four cards are an Ace) = 1 / 270725

P( All four cards are an Ace) = 0.0000037