{"id":7028,"date":"2021-10-22T15:40:59","date_gmt":"2021-10-22T10:10:59","guid":{"rendered":"https:\/\/mathemerize.com\/?p=7028"},"modified":"2021-10-25T01:43:37","modified_gmt":"2021-10-24T20:13:37","slug":"a-fair-die-is-tossed-eight-times-the-probability-that-a-third-six-is-observed-on-the-eight-throw-is","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/a-fair-die-is-tossed-eight-times-the-probability-that-a-third-six-is-observed-on-the-eight-throw-is\/","title":{"rendered":"A fair die is tossed eight times. The probability that a third six is observed on the eight throw, is"},"content":{"rendered":"

Solution :<\/h2>\n

Probability of getting success, p = \\(1\\over 6\\)<\/p>\n

and probability of failure, q = \\(5\\over 6\\)<\/p>\n

Now, we must get 2 sixes in seven throws, so probability is \\(^7C_2\\)\\(({1\\over 6})^2\\)\\(({5\\over 6})^5\\)<\/p>\n

and the probability that 8th throw is \\(1\\over 6\\).<\/p>\n

\\(\\therefore\\)\u00a0 \u00a0Required Probability = \\(^7C_2\\)\\(({1\\over 6})^2\\)\\(({5\\over 6})^5\\)\\(1\\over 6\\)<\/p>\n

= \\(^7C_2\\times 5^5\\over {6^8}\\)<\/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

Consider rolling two dice once. Calculate the probability of rolling a sum of 5 or an even sum.<\/a><\/p>\n

If A and B are two mutually exclusive events, then<\/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

A problem in mathematics is given to three students A, B, C and their respective probability of solving the problem is \\(1\\over 2\\), \\(1\\over 3\\) and \\(1\\over 4\\). Probability that the problem is solved is<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"

Solution : Probability of getting success, p = \\(1\\over 6\\) and probability of failure, q = \\(5\\over 6\\) Now, we must get 2 sixes in seven throws, so probability is \\(^7C_2\\)\\(({1\\over 6})^2\\)\\(({5\\over 6})^5\\) and the probability that 8th throw is \\(1\\over 6\\). \\(\\therefore\\)\u00a0 \u00a0Required Probability = \\(^7C_2\\)\\(({1\\over 6})^2\\)\\(({5\\over 6})^5\\)\\(1\\over 6\\) = \\(^7C_2\\times 5^5\\over {6^8}\\) Similar …<\/p>\n

A fair die is tossed eight times. The probability that a third six is observed on the eight throw, is<\/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":"\nA fair die is tossed eight times. The probability that a third six is observed on the eight throw, is<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mathemerize.com\/a-fair-die-is-tossed-eight-times-the-probability-that-a-third-six-is-observed-on-the-eight-throw-is\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"A fair die is tossed eight times. The probability that a third six is observed on the eight throw, is\" \/>\n<meta property=\"og:description\" content=\"Solution : Probability of getting success, p = (1over 6) and probability of failure, q = (5over 6) Now, we must get 2 sixes in seven throws, so probability is (^7C_2)(({1over 6})^2)(({5over 6})^5) and the probability that 8th throw is (1over 6). (therefore)\u00a0 \u00a0Required Probability = (^7C_2)(({1over 6})^2)(({5over 6})^5)(1over 6) = (^7C_2times 5^5over {6^8}) Similar … A fair die is tossed eight times. 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