Probability

Question

In a standard deck of cards, what is the probability of selecting  (assume no replacement) 

 

a) a spade or club

b) a diamond or queen

c) a spade and then a club

d) a pair of 2's

Answer 1

standard deck has 52 cards, half black, half red

1/4 clubs 1/4 spades 1/4 diamonds 1/4 hearts

1/13 twos, 1/13 queens

a) half the cards are black

so probability of spade or club is half = 1/2 = .5 = 50%

b) 1/4 are diamonds, 1/13 are queens 1/52 is the one & only diamond queen

P(diamond or queen) = P(d) + P(q) - P(d^q) = (1/4) + (1/13) - (1/52) = (13 +4 -1)/52 = 16/52= 4/13 = about 30.8%

c) Probability of spade then club = P(s then c) = 1/4 x 13/51 = 13/204= about 6.37%

d) 6 ways to get 2 twos

2 of spades & 2 of clubs

2 of spades & 2 of diamonds

2 of spades & 2 of hearts

2 of clubs & 2 of diamonds

2 of clubs & 2 of hearts

2 of diamonds & 2 of hearts

total ways of randomly selecting 2 cards is 52C2 = 52!/2!50!= 52(51)/2= 26(51) = 1326

Probability of 2 two's = 6/1326= 1/221 = about .45%

Answer 2

There are 52 cards in a deck.  If you draw a card and don't replace it, there are only 51 cards left

 

a)  There are 13  spades and 13 clubs

 

     P(spade or club) = (13+13)/52 = 26/52 = 1/2 = 0.5

 

b)  There are 13 diamonds and 4 queens; but one is the Q of diamonds which we've already counted.  So there are only three additional queens:

 

     P(diamond or queen) = (13 + 3)/52 = 16/52 = 4/13 ≈ 0.308

 

c) Here we draw two cards, a spade then a club:

 

     P(spade and club) = 13/52 * 13/51 = 169/2652 ≈ 0.064

 

d)  There are 4 twos in the deck.  If you pick one and don't replace it, there are only 3 twos left (and 51 cards):

 

     P(two and two) = 4/52 * 3/51 = 1/221 ≈ 0.0045