{"id":6956,"date":"2021-10-22T02:04:26","date_gmt":"2021-10-21T20:34:26","guid":{"rendered":"https:\/\/mathemerize.com\/?p=6956"},"modified":"2021-10-25T01:59:49","modified_gmt":"2021-10-24T20:29:49","slug":"let-a-and-b-be-two-events-such-that-pa-cup-b-1-6-pa-cap-b-1-4-and-pa-1-4-where-a-stands-for-complement-of-a-then-prove-that-events-a-and-b-independent","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/let-a-and-b-be-two-events-such-that-pa-cup-b-1-6-pa-cap-b-1-4-and-pa-1-4-where-a-stands-for-complement-of-a-then-prove-that-events-a-and-b-independent\/","title":{"rendered":"Let A and B be two events such that P(A \\(\\cup\\) B)’ = 1\/6, P(A \\(\\cap\\) B) = 1\/4 and P(A)’ = 1\/4 where A’ stands for complement of A. Then prove that events A and B independent"},"content":{"rendered":"

Solution :<\/h2>\n

Given P(A \\(\\cup\\) B)’ = 1\/6, P(A \\(\\cap\\) B) = 1\/4 and P(A)’ = 1\/4<\/p>\n

\\(\\therefore\\)\u00a0 \u00a0P(A \\(\\cup\\) B) = 1 – P(A \\(\\cup\\) B)’<\/p>\n

= 1 – \\(1\\over 6\\) = \\(5\\over 6\\)<\/p>\n

and P(A) = 1 – P(A)’ = 1 – \\(1\\over 4\\) = \\(3\\over 4\\)<\/p>\n

P(A \\(\\cup\\) B) = P(A) + P(B) – P(A \\(\\cap\\) B)<\/p>\n

\\(\\implies\\) \\(5\\over 6\\) = \\(3\\over 4\\) + P(B) – \\(1\\over 4\\)<\/p>\n

P(B) = \\(1\\over 3\\)<\/p>\n

\\(\\implies\\) A and B are not equally likely,<\/p>\n

Also, P(A \\(\\cap\\) B) = P(A).P(B) = \\(1\\over 4\\)<\/p>\n

So, events are independent.<\/p>\n

\n

Similar Questions<\/h3>\n

The probability of India winning a test match against the west indies is 1\/2 assuming independence from match to match. The probability that in a match series India\u2019s second win occurs at the third test 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

Three groups A, B, C are contesting for positions on the board of directors of a company. The probabilities of their winning are 0.5, 0.3, 0.2 respectively. If the group A wins, the probability of introducing a new product is 0.7 and the corresponding probabilities for group B and C are 0.6 and 0.5 respectively. Find the probability that the new product will be introduced.<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"

Solution : Given P(A \\(\\cup\\) B)’ = 1\/6, P(A \\(\\cap\\) B) = 1\/4 and P(A)’ = 1\/4 \\(\\therefore\\)\u00a0 \u00a0P(A \\(\\cup\\) B) = 1 – P(A \\(\\cup\\) B)’ = 1 – \\(1\\over 6\\) = \\(5\\over 6\\) and P(A) = 1 – P(A)’ = 1 – \\(1\\over 4\\) = \\(3\\over 4\\) P(A \\(\\cup\\) B) = P(A) + …<\/p>\n

Let A and B be two events such that P(A \\(\\cup\\) B)’ = 1\/6, P(A \\(\\cap\\) B) = 1\/4 and P(A)’ = 1\/4 where A’ stands for complement of A. 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