In: Statistics and Probability
Suppose you shuffle a deck of cards and set it down on the table. (a) What is the probability that the top ten cards contain no aces? (There are four aces in the whole deck). (b) What is the probability that the top four cards are all the same suit? (There are four suits, and thirteen cards belonging to each suit). (c) What is the probability that the top four cards contain (exactly one) pair of the same card? (There are thirteen different cards and each occurs four times).
Given, there are a deck of cards, 52 cards- 4 different suits, each suit containing 13 different cards
ANSWER a:
Probability that the top ten cards contain NO ACES = ( 48C10 / 52C10 )
= ( 48 x 47 x 46 x 45 x 44 x 43 x 42 x 41 x 40 x 39 ) / ( 52 x 51 x 50 x 49 x 48 x 47 x 46 x 45 x 44 x 43 )
= 246/595
( SINCE, 48C10 - In 52 cards, 4 Aces are there. So, 52-4 = 48 )
ANSWER b:
The required probability = ( 4 x 13C4 ) / 52C4 = ( 4 x 10 x 11 x 12 x 13) / ( 52 x 51 x 50 x 49 ) = 44/4165
( Since, there are 4 suits, that one suit could be any of the 4.
13C4 - the 4 cards of any given suit can be chosen in 13C4 ways)
ANSWER c:
The required probability= ( 13 x 4C2 x 48C2 ) / 52C4 = ( 13 x 6 x 48 x 47 x 4! ) / ( 2 x 49 x 50 x 51 x 52 ) = 6768/20825
( SINCE, there are 13 different numbers
the 2 numbers can be from any of the 4 suits, i.e; 4C2
the remaining 2 numbers can be any 2 of the (52-4)=48 cards, 48C2)