Finding the probability of P((A ∪ B) ∪ C)?
Consider the random experiment in which one card is drawn from a standard deck of 52. Let RED be the event that a red card (hearts or diamonds) is drawn, BLACK be the event that a black card (clubs or spades) is drawn, EVEN means an even-numbered card is drawn (2, 4, 6, 8 or 10); ODD means an odd-numbered card is drawn (3, 5, 7 or 9 - aces not included); COURT means a court card (jack, queen, king or ace); TWO means a 2 of any suit is drawn, etc. Calculate the probability of:
(i) P((ODD ∪ BLACK) ∪ EVEN)
- PuzzlingLv 74 weeks agoFavourite answer
There are 5 even cards in each of the 4 suits (20).
There are 4 odd cards in each of the 4 suits (16).
There are 26 black cards.
If we add that all up we get:
20 + 16 + 26 = 62
Clearly we've double-counted because there are 10 cards that are black & even and 8 cards that are black and odd.
Subtracting that we get:
62 - 18
This makes sense because the only cards that aren't included are the red court cards (4 cards × 2 red suits = 8). So another way to count it would be:
52 - 8
Dividing by the total number of cards, the probability is: