p(A \u222a B) =p(A) + p(B) \u2013 p(A \u2229 B)
\nThis rule is also called the inclusion-exclusion principle of probability.<\/p>\n
This formula reduces to
\np(A \u222a B) =p(A) + p(B)
\nif A and B are mutually exclusive, since p(A \u2229 B) = 0 in such a case.<\/p>\n
Rule 2:<\/strong><\/p>\n p(A \u2229 B) =p(A) * p(B\/A) = p(B) * p(A\/B)<\/p>\n where p(A\/B) represents the conditional probability of A given B and p(B\/A) represents the conditional probability of B given A.<\/p>\n (a) p(A) and p(B) are called the marginal probabilities of A and B respectively. This rule is also called as the multiplication rule of probability.<\/p>\n (b) p(A \u2229 B) is called the joint probability of A and B.<\/p>\n (c) If A and B are independent independent events, this formula reduces to<\/p>\n p(A \u2229 B)=p(A) * p(B).<\/p>\n since when A and B are independent<\/p>\n p(A\/B) = p(A)<\/p>\n and p(B\/A) = p(B)<\/p>\n i.e. the conditional probabilities become same as the marginal (unconditional) probabilities.<\/p>\n (d) If A and B are independent, then so are A and BC; AC and B and AC and BC.<\/p>\n (e) Condition for three events to independent:<\/p>\n Events A, B and C are independent iff<\/p>\n p(ABC)=p(A) p(B) p(C) Note:<\/strong> If A, B, C are independent, then A will be independent of any event formed from B and C.<\/p>\n For instance, A is independent of B \u222a C.<\/p>\n Rule 3: Complementary Probability<\/strong><\/p>\n p(A) = 1 \u2013 p(AC)<\/p>\n p(AC ) is called the complementary probability of A and p(AC) represents the probability that the event A will not happen.<\/p>\n p(A) = 1 \u2013 p(AC)<\/p>\n p(AC) is also written as p(A\u2032)<\/p>\n Notice that<\/p>\n p(A) + p(A\u2032) =1 Also notice that by De Morgan\u2019s law since<\/p>\n AC \u2229 BC = (A \u222a B)C Rule 4: Conditional Probability Rule<\/strong><\/p>\n Starting from the multiplication rule p(A\/B) = p(A \u2229 B)\/p(B)<\/p>\n By interchanging A and B in this formula we get<\/p>\n p(B\/A) = p(A \u2229 B)\/p(A)<\/p>\n Rule 5: Rule of Total Probability<\/strong><\/p>\n Consider an event E which occurs via two different events A and B. Further more, let A and B be mutually Now, the probability of E is given by value of total probability as<\/p>\n P(E) =P(A \u2229 E) + P(B \u2229 E) = P(A) * P(E\/A) + P(B) *(E\/B)<\/p>\n This is called rule of total probability. Rule 6: Baye\u2019s theorem<\/strong><\/p>\n If E1, E2, E3 …… EN are \u2018n\u2019 mutually exclusive events in a sample space \u2018S\u2019 such that S = E1 \u222a E2 \u222a E3 …….EN. There are six rules of probability using which the probability of any compound event involving arbitrary events A and B,<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1754,5],"tags":[1773,1772,1771],"class_list":["post-6049","post","type-post","status-publish","format-standard","hentry","category-probability-and-statistics","category-ee","tag-bayes-theorem","tag-complementary-probability","tag-conditional-probability-rule"],"_links":{"self":[{"href":"https:\/\/www.madeeasy.in\/study\/wp-json\/wp\/v2\/posts\/6049","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.madeeasy.in\/study\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.madeeasy.in\/study\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.madeeasy.in\/study\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.madeeasy.in\/study\/wp-json\/wp\/v2\/comments?post=6049"}],"version-history":[{"count":0,"href":"https:\/\/www.madeeasy.in\/study\/wp-json\/wp\/v2\/posts\/6049\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.madeeasy.in\/study\/wp-json\/wp\/v2\/media?parent=6049"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.madeeasy.in\/study\/wp-json\/wp\/v2\/categories?post=6049"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.madeeasy.in\/study\/wp-json\/wp\/v2\/tags?post=6049"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\np(AB)=p(A) p(B)
\np(AC)=p(A) p(C)\u00a0 A, B, C are pairwise independent
\np(BC)=p(B) p(C)<\/p>\n
\ni.e. A and A\u2032 are mutually exclusion as well as collectively exhaustive.<\/p>\n
\np(AC \u2229 BC) =p(A \u222a B)C = 1 \u2013 p(A \u222a B)
\ni.e.\u00a0 p(neither A nor B) = 1 \u2013 p(either A or B)<\/p>\n
\np(A \u2229 B) =p(B) * p(A\/B)
\nby cross multiplying we get the conditional probability formula<\/p>\n
\nexclusive and collectively exhaustive events. This situation may be represented by following tree diagram<\/p>\n
\nSometimes however, we may wish to know that, given that the event E has already occurred, what is the probability that it occurred with A? In this case we can use Bayes Theorem given below.<\/p>\n
\nIf A is any arbitrary event in sample space then, probability of even Ei when A has already occurred i.e.,<\/p>\n![\"Baye\u2019s](\"https:\/\/study.madeeasy.in\/wp-content\/uploads\/2025\/08\/baye-theorem.jpg\")
<\/p>\n","protected":false},"excerpt":{"rendered":"