If events A and B are mutually exclusive, then P (A ∪ B) = P (A) + P (B)
Probability of an event A not happening = P(Ac) = 1 - P(A)
If A and B are Independent Events, then P(A ∩ B) = P(A) * P(B)
Probability of A given that B has occurred is -
In case of experiments with only 2 outcomes possible (tossing a coin, passing or failing, hitting or not hitting a target etc.), the probability of getting r successes in n trials (n ≥ r) is nCr * pr * qn-r, where p is the probability of success and q is the probability of failure.
P(A∪B∪C) = P(A) + P(B) + P(C) - P(A ∩ B) - P(B∩C) - P(A ∩ C) + P(A ∩ B ∩ C).
P(A ∩ B) = P(A) * P(B), if A and B are independent events.
P(A ∩ B) = 0, if A and B are mutually exclusive events.
P(A) + P(AC) = 1.
P(AC∩ BC) = P[(A ∪ B)C] = 1 - P(A ∪ B).
P(AC ∪ BC) = P[(A ∩ B) C] = 1 - P(A ∩ B).
If probability of an event in defined as the odds against or odds in favour then the logic to solve this is as follows. If the odd in favour of happening of an event are given to be in the ratio of p : q. Then the probability then the event will happen is P/(p+q) .
If the odds against happening of an event is given to be in the ratio of r : s, then the probability that the event will happen is S/(r+s) . Hence it is to be understood clearly what is being given in the question.
Deck of Cards
A pack of cards has 52 cards which are divided into four categories: Spades, Clubs, Hearts and Diamonds.
Each category of cards has 13 cards in all out of which 9 cards are numbered 2 to 10 and other 4 cards are face cards with alphabets A, J, Q and K.
A stands for Ace; J stands for Jack; Q stands for Queen and K for King.
Coin
When we toss a fair coin, the outcome will always be either Head (H) or Tail (T).
In n throws of coin, total number of possible cases will be 2n.
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