Hence machine will be working if less than three components fail. In a town, 80% of all the families own a television set. The probability that 7 families have television (X = 7): ∴ P(X = 7) = 120 x (0.8)7 (0.2)3 = 0.2013. In this case number of trials = n = 14, Probability that component fails (success) = 0.1. The probability that all will develop immunity (X = 8): A machine has fourteen identical components that function independently. The prefix “bi” means two. In this article, we shall study to solve problems of probability based on the concept of the binomial distribution. The probability that machine is working (X < 3): ∴ P(X < 3) = P(X = 0) + P(X = 1) + P(x = 2), ∴ P(X < 3) = 14C0 (0.1)0 (0.9)14 – 0 + 14C1 (0.1)1 (0.9)14 – 1 + 14C2 (0.1)2 (0.9)14 – 2, ∴ P(X < 3) = 1x 1 x  (0.9)14 + 14 x  (0.1)1 (0.9)13 + 91 x  (0.1)2 (0.9)12, ∴ P(X < 3) = (1 x  (0.9)2 + 14 x  0.1 x (0.9) + 91 x  (0.1)2 )(0.9)12, ∴ P(X < 3) = (0.81 + 1.26 + 0.91 )(0.9)12. In simple words, a binomial distribution is the probability of a success or failure results in an experiment that is repeated a few or many times. The probability that a bomb will hit a target is 0.8. ∴ P(X = 5) =  126 x (1/2)9 = 126 x (1/512) = 63/256 = 0.2461. The probability that two third support ammendment (X ≥ 6): ∴ P(X ≥ 6) = P(X = 6) + P(X = 7) + P(x = 8) + P(x = 9), ∴ P(X ≥ 6) = 9C6 (0.8)6 (0.2)9 – 6 + 9C7 (0.8)7 (0.2)9 – 7 + 9C8 (0.8)8 (0.2)9 – 8 + 9C9 (0.8)9 (0.2)9 – 9, ∴ P(X ≥ 6) = 84 x (0.8)6 (0.2)3 + 36 x (0.8)7 (0.2)2 + 9 x (0.8)8 (0.2)1 + 1x (0.8)9 (0.2)0, ∴ P(X ≥ 6) = 84 x (0.8)6 (0.2)3 + 36 x (0.8)7 (0.2)2 + 9 x (0.8)8 (0.2)1 + 1x (0.8)9 x 1, For More Topics in Probability Click Here, For More Topics in Mathematics Click Here, Your email address will not be published. If 10 families are interviewed at random, find the probability that a) seven families own a television set b) atmost three families own a television set. Find the probability that the machine will be working. Science > Mathematics > Statistics and Probability > Probability > Binomial Distribution. The probability of getting atmost 3 correct answers (X ≤ 3): ∴ P(X ≤ 3) = P(X = 0) + P(X = 1)  + P(X = 2) + P(X = 3), ∴ P(X ≤ 3) = 5C0 (1/4)0 (3/4)5 – 0 + 5C1 (1/4)1 (3/4)5 – 1+ 5C2 (1/4)2 (3/4)5 – 2 +  5C3 (1/4)3 (3/4)5 – 3, ∴ P(X ≤ 3) = 1 x 1 x  (3/4)5+ 5 x  (1/4)1 (3/4)4  + 10 x (1/4)2 (3/4)3 + 10 x (1/4)3 (3/4)2, ∴ P(X ≤ 3) = (243/1024) + 5 x  (1/4) x (81/256) + 10 x (1/16) (27/64) + 10 x (1/64) (9/16), ∴ P(X ≤ 3) = (243/1024) + (405/1024) + (270/1024) + (90/1024). The probability of gettingin head in first four tosses and tails in last five tosses : ∴ P(X) = (1/2)(1/2)(1/2)(1/2)(1/2)(1/2)(1/2)(1/2)(1/2)(1/2) = 1/512 = 0.001953, Ans: The probability of getting exactly 5 heads is 63/256 or 0.2461 and the probability of getting head in the first four tosses and tails in the last five tosses is 1/512 or 0.001953. The probability of getting exactly 3 answers correct (X = 3): ∴ P(X = 3) = 10 x  (1/64) (9/16) = 90/1024 = 45/512 = 0.0879. In binomial probability distribution, the number of ‘Success’ in a sequence of n experiments, where each time a question is asked for yes-no, then the boolean-valued outcome is represented either with success/yes/true/one (probability p) or failure/no/false/zero (probability q = 1 − p). Example – 01: An unbiased coin is tossed 5 times. In this article, we shall study to solve problems of probability based on the concept of the binomial distribution. These are also known as Bernoulli trials and thus a Binomial distribution is the result of a sequence of Bernoulli trials. What is the probability that at least two-thirds of them will support the amendment? The random variable X is the number of questions answer correctly. The probability that exactly 4 will develop immunity (X = 4): ∴ P(X = 4) = 70 x (0.8)4 (0.2)4  = 0.04587. Every trial has a possible result, selected from S (for success), F (for failure), and each trial’s probability would be the same. An unbiased coin is tossed 9 times. The probability of getting at least 4 heads (X ≥ 4): ∴ P(X ≥ 4) = 5C4 (1/2)4 (1/2)5 – 4 + 5C5 (1/2)5 (1/2)5 – 5, ∴ P(X ≥ 4) = 5 x (1/2)4 (1/2)1 + 1 x  (1/2)5 (1/2)0, ∴ P(X ≥ 4) = 5 x (1/32) + 1 x  (1/32) = 6/32 = 3/16 = 0.1875, Ans: The probability of getting exactly 3 heads is 5/16 or 0.3125 and the probability of getting at least 4 heads is 3/16 or 0.1875.

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