Probability
Independent Events

If A and B are independent events, then
P(A ∩ B) = P(A) P(B)
\tt P(A \cup B) = 1  P(\overline{A}) P(\overline{B})  If A mapping is selected at random from the set of mappings from A to B, then the probability that the mapping is oneone function = \frac{{^{n(B)}P_{n(A)}}}{n(B)^{n(A)}}
 If a mapping is selected at random from the set of mappings from A to B, then the probability that the mapping is many one function is = 1 \frac{{^{n(B)}P_{n(A)}}}{n(B)^{n(A)}}
 If a mapping is selected at random from the set of mappings from A to B, then the probability that the mapping is constant function is = \frac{n(B)}{n(B)^{n(A)}}
 If a mapping is selected at random from the set of mapping from A to B, then the probability that the mapping is onto function is = \frac{\sum_{r = 1}^{n}(1)^{nr} \ {^nc_r}r^{m}}{n(B)^{n(A)}} where n(A) = m, n(B) = n
 A mapping is selected at random from the set of mappings from A to B, then the probability that the mapping is bijection is = \frac{n(A)!}{n(B)^{n(A)}} \ (or) \ \frac{n(B)!}{n(B)^{n(A)}}
 The probability that the 'r' shoes are selected at random from the n pair of shoes if r shoes does not contain any pair is = \frac{{^nc_r}2^{r}}{{^{2n}c_r}}
 The probability that the 'r' shoes are selected at random from the n pair of shoes if 'r' shoes contain at least one pair is = 1  \frac{{^nc_r}2^{r}}{{^{2n}c_r}}
 The probability that the 'r' shoes are selected at random from the n pair of shoes if 'r' shoes contain exactly k pairs is \frac{{^nc_k}\ ^{nk}c_{r2k}\ 2^{r2k}}{{^{2n}c_r}}
 If p and q are the probability of success and failure of a game in which A, B and C play in order, then probability of A wins = \frac{p}{1  q^{2}} = \frac{1}{1 + q + q^{2}}
 If p and q are the probability of success and failure of a game in which A, B and C play in order, then probability of B wins = \frac{qp}{1  q^{3}} = \frac{q}{1 + q + q^{2}}
 If p an q are the probability of success and failure of a game in which A, B and C play in order, then probability of C wins \frac{q^{2}p}{1  q^{3}} = \frac{q^{2}}{1 + q + q^{2}}
 If p and q are the probability of success and failure of a game in which A, B and C play in order, then the ratio of their success = 1 : q : q^{2}
 If p and q are the probabilities of the success and failure of the game in which A and B play and A starts the game, then the probability of A's win = \frac{p}{1  q^{2}} = \frac{1}{1 + q}
 If p and q are the probability of the success and failure of the game in which A and B play and A starts the game, then the probability of B's win = \frac{qp}{1  q^{2}} = \frac{q}{1 + q}
 If p and q are the probability of the success and failure of the game in which A and B play and A starts the game, then the probability of ratio of their success is = 1 : 2
 If p_{A} and q_{A} are the probabilities of success and failure of A and p_{B} and q_{B} are the probability of success and failure of B and if A starts the game, then probability of A's win = \frac{p_{A}}{1  q_{A}q_{B}}
 A man has a bunch of n keys of which only one fits the door. He tries them successively without repetition, then the probability that the door is opened at any trial (less than n) = \frac{1}{n}
 E and F are mutually exclusive events and the random experiment is repeated till E or F occurs. Then the probability that E occurs before F is = \frac{P(E)}{P(E) + P(F)}
 If A has m shares in a lottery where there are m prizes and n blanks, then the probability of A is winning = 1  \frac{^nc_m}{^{(m + n)}c_m}
 Using Vertices of a polygon having n sides a triangle is constructed at random. Then the probability that the triangle so formed is such that no side of the polygon is side of the triangle is = \frac{(n  4)(n  5)}{(n  1)(n  2)}
 Out of n persons sitting at a round table, there persons are selected at random, then the probability that no two of them are consecutive is = \frac{(n  4)(n  5)}{(n  1)(n  2)}
 If r (1 ≤ r ≤ 7) squares are selected at random from a chess board, then the probability that they lie on a diagonal is = \frac{4\left[^{7}c_r + {^6}c_r + {^5}c_r + .... + {^r}c_r\right] + 2\left[{^8}c_r\right]}{^{64}c_r}
 If n whole numbers are taken at random and multiplied together, than the probability that the last digit of the product is either 1, 3, 7 (or) 9 is = \left(\frac{2}{5}\right)^{n}
 If n whole numbers are taken at random and multiplied together, then the probability that the last digit of the product is either 2, 4, 6 (or) 8 is = \frac{4^{n}  2^{n}}{5^{n}}
 If n whole numbers are taken at random and multiplied together, then the probability that the last digit of the product is 5 = \frac{5^{n}  4^{n}}{10^{n}}
 If n whole numbers are taken at random and multiplied together, then the probability that the last digit of the product is 0 is = \frac{10^{n}  8^{n}  5^{n} + 4^{n}}{10^{n}}
View the Topic in this video From 44:00 To 57:28
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1. Let A and B be two events associated with the same random experiment, then E and F are said to be independent if
P(A ∩ B) = P(A) . P(B)
2. Three events A, B and C are said to be mutually independent, if
P(A ∩ B) = P(A) P(B)
P(A ∩ C) = P(A) P(C)
P(B ∩ C) = P(B) P(C)
and P(A ∩ B ∩ C) = P(A) P(B) P(C)