Answer:
Step-by-step explanation:
Given that an ordinary deck of 52 cards is shuffled and the cards are then turned over one at a time until the first ace appears.
the first ace is the 20th card to appear
This implies that there are still 32 cards with 3 aces in it.
But 2 clubs can be either 1 remaining or nothing remaining
a) ace of spades
3 aces out of 32
Prob = [tex]\frac{3}{32}[/tex]
b) 2 of clubs
P (2 club already gone in first 19 cards) +P(2 club is in the remaining)
= 0+P(first 20 cards not 2 clubs )*P(2 club in 21st draw)
= [tex](\frac{51}{52} )(\frac{1}{32}) \\\\= \frac{51}{1664}[/tex]
