{"id":6972,"date":"2021-10-22T02:24:46","date_gmt":"2021-10-21T20:54:46","guid":{"rendered":"https:\/\/mathemerize.com\/?p=6972"},"modified":"2021-10-25T01:57:45","modified_gmt":"2021-10-24T20:27:45","slug":"let-a-b-and-c-are-pairwise-independent-events-with-pc-0-and-pa-cap-bcap-c-0-then-pa-cap-b-c-is-equal-to","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/let-a-b-and-c-are-pairwise-independent-events-with-pc-0-and-pa-cap-bcap-c-0-then-pa-cap-b-c-is-equal-to\/","title":{"rendered":"Let A, B and C are pairwise independent events with P(C) > 0 and \\(P(A \\cap B\\cap C)\\) = 0. Then \\(P(A’ \\cap B’\/C)\\) is equal to"},"content":{"rendered":"

Solution :<\/h2>\n

\\(P({A’ \\cap B’\\over C})\\) = \\(P(A’ \\cap B’ \\cap C)\\over P(C)\\)<\/p>\n

= \\(P(C) – P(A \\cap C) – P(B \\cap C) + P(A \\cap B\\cap C)\\over P(C)\\)\u00a0 \u00a0……..(i)<\/p>\n

Given, \\(P(A \\cap B\\cap C)\\)\u00a0 = 0 and A, B and C are pairwise independent.<\/p>\n

\\(\\therefore\\)\u00a0 \\(P(A \\cap C)\\) = P(A).P(C)<\/p>\n

and \\(P(B \\cap C)\\) = P(B).P(C)<\/p>\n

\\(\\therefore\\)\u00a0 \\(P({A’ \\cap B’\\over C})\\) = \\(P(C) – P(A).P(C) – P(B).P(C) + 0\\over P(C)\\)<\/p>\n

= 1 – P(A) – P(B) = P(A’) – P(B)<\/p>\n

\n

Similar Questions<\/h3>\n

A multiple choice examination has 5 questions. Each question has three alternative answers of which exactly one is correct. The probability that a student will get 4 or more correct answers just by guessing is<\/a><\/p>\n

Two aeroplanes I and II bomb a target in succession. The probabilities of I and II scoring a hit correctly are 0.3 and 0.2, respectively. The second plane will bomb only if the first misses the target. The probability that the target is hit by the second plane, is<\/a><\/p>\n

A pair of fair dice is thrown independently three times. The probability of getting a score of exactly 9 twice is<\/a><\/p>\n

A and B play a game, where each is asked to select a number from 1 to 25. If the two numbers match, both of them win a prize. The probability that they will not win a prize in a single trial is<\/a><\/p>\n

One ticket is selected at random from 50 tickets numbered 00, 01, 02, \u2026\u2026, 49. Then, the probability that the sum of the digits on the selected ticket is 8, given that the product of these digits is zero equal to<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"

Solution : \\(P({A’ \\cap B’\\over C})\\) = \\(P(A’ \\cap B’ \\cap C)\\over P(C)\\) = \\(P(C) – P(A \\cap C) – P(B \\cap C) + P(A \\cap B\\cap C)\\over P(C)\\)\u00a0 \u00a0……..(i) Given, \\(P(A \\cap B\\cap C)\\)\u00a0 = 0 and A, B and C are pairwise independent. \\(\\therefore\\)\u00a0 \\(P(A \\cap C)\\) = P(A).P(C) and \\(P(B \\cap C)\\) …<\/p>\n

Let A, B and C are pairwise independent events with P(C) > 0 and \\(P(A \\cap B\\cap C)\\) = 0. Then \\(P(A’ \\cap B’\/C)\\) is equal to<\/span> Read More »<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-global-header-display":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":""},"categories":[43,44],"tags":[],"yoast_head":"\nLet A, B and C are pairwise independent events with P(C) > 0 and \\(P(A \\cap B\\cap C)\\) = 0. 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