Total number of ways of dividing 48 cards(Excluding 4 Aces) in 4 groups = \(48!\over (12!)^4 4!\)
Now, distribute exactly one Ace to each group of 12 cards. Total number of ways = \(48!\over (12!)^4 4!\) \(\times\) 4!
Now, distribute these groups of cards among four players
= \(48!\over (12!)^4 4!\) \(\times\) 4!4!
= \(48!\over (12!)^4\) \(\times\) 4!
The probability of India winning a test match against the west indies is 1/2 assuming independence from match to match. The probability that in a match series India’s second win occurs at the third test is
Three houses are available in a locality. Three persons apply for the houses. Each applies for one house without consulting others. The probability that three apply for the same house is
If A and B are two mutually exclusive events, then
A coin is tossed successively three times. Find the probability of getting exactly one head or two heads.
A die is thrown. Let A be the event that the number obtained is greater than 3. Let B be the event that the number obtained is less than 5. Then, \(P(A \cup B)\) is