{"id":6963,"date":"2021-10-22T02:14:14","date_gmt":"2021-10-21T20:44:14","guid":{"rendered":"https:\/\/mathemerize.com\/?p=6963"},"modified":"2021-10-25T01:59:54","modified_gmt":"2021-10-24T20:29:54","slug":"consider-5-independent-bernoullis-trials-each-with-probability-of-success-p-if-the-probability-of-atleast-one-failure-is-greater-than-or-equal-to-31over-32-then-p-lies-in-the-interval","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/consider-5-independent-bernoullis-trials-each-with-probability-of-success-p-if-the-probability-of-atleast-one-failure-is-greater-than-or-equal-to-31over-32-then-p-lies-in-the-interval\/","title":{"rendered":"Consider 5 independent Bernoulli’s trials each with probability of success p. If the probability of atleast one failure is greater than or equal to \\(31\\over 32\\), then p lies in the interval"},"content":{"rendered":"

Solution :<\/h2>\n

Here, n = 5 and r \\(\\ge\\) 1<\/p>\n

\\(\\therefore\\)\u00a0 \u00a0p(X = r) = \\(^nC_r\\) \\(p^{n-r}\\) \\(q^r\\)<\/p>\n

P(X \\(\\ge\\) 1) = 1 – P(X = 0)<\/p>\n

= 1 – \\(^5C_0 . p^5 . q^0\\) \\(\\ge\\) \\(31\\over 32\\)\u00a0 \u00a0[Given]<\/p>\n

\\(\\implies\\)\u00a0 \u00a0\\(p^5\\) \\(\\le\\) 1 – \\(31\\over 32\\) = \\(1\\over 32\\)<\/p>\n

\\(\\therefore\\)\u00a0 p \\(\\le\\) \\(1\\over 2\\) and p \\(\\ge\\) 0<\/p>\n

\\(\\implies\\)\u00a0 p \\(\\in\\)\u00a0 [0, 1\/2]<\/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

Let A and B be two events such that P(A \\(\\cup\\) B)\u2019 = 1\/6, P(A \\(\\cap\\) B) = 1\/4 and P(A)\u2019 = 1\/4 where A\u2019 stands for complement of A. Then prove that events A and B independent<\/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 : Here, n = 5 and r \\(\\ge\\) 1 \\(\\therefore\\)\u00a0 \u00a0p(X = r) = \\(^nC_r\\) \\(p^{n-r}\\) \\(q^r\\) P(X \\(\\ge\\) 1) = 1 – P(X = 0) = 1 – \\(^5C_0 . p^5 . q^0\\) \\(\\ge\\) \\(31\\over 32\\)\u00a0 \u00a0[Given] \\(\\implies\\)\u00a0 \u00a0\\(p^5\\) \\(\\le\\) 1 – \\(31\\over 32\\) = \\(1\\over 32\\) \\(\\therefore\\)\u00a0 p \\(\\le\\) \\(1\\over 2\\) and …<\/p>\n

Consider 5 independent Bernoulli’s trials each with probability of success p. If the probability of atleast one failure is greater than or equal to \\(31\\over 32\\), then p lies in the interval<\/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":"\nConsider 5 independent Bernoulli's trials each with probability of success p. 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