These are also known as Bernoulli trials and thus a Binomial distribution is the result of a sequence of Bernoulli trials. 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 5 lines busy (X = 5): ∴ P(X = 5) = 252 x (0.2 x 0.8)5 = 252 x (0.16)5 = 0.0264. The probability that a bomb will hit a target is 0.8. The random variable X is the number of questions answer correctly. The probability that half of them will recover is 0.1852. The probability of getting more heads than tail (X ≥ 5): ∴ P(X ≥ 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8), ∴ P(X ≥ 5) = 8C5 (1/2)5 (1/2)8 – 5 + 8C6 (1/2)6 (1/2)8 – 6 + 8C7 (1/2)7 (1/2)8 – 7 + 8C8 (1/2)8 (1/2)8 – 8, ∴ P(X ≥ 5) = 56 x (1/2)5 (1/2)3 + 28 x (1/2)6 (1/2)2 + 8 x (1/2)7 (1/2)1 + 1 x (1/2)8 (1/2)0, ∴ P(X ≥ 5) = 56 x (1/2)8 + 28 x (1/2)8 + 8 x (1/2)8 + 1 x (1/2)8, ∴ P(X ≥ 5) = (56 + 28 + 8 + 1)x (1/256) = 93/256 = 0.3633. The probability that all will develop immunity (X = 8): A machine has fourteen identical components that function independently. The probability that a person who undergoes a kidney operation will recover is 0.7, Find the probability that of the six patients who undergo similar operations: a) none will recover, b) all will recover, c) half of them will recover and iv) aleast half will recover. What is an Experiment? What is the probability that at least two-thirds of them will support the amendment? c) at least one correct answer. Find the probability that out of 10 bombs dropped, exactly two will miss the target. Machine will stop working if three or more components fail. In this case number of lines = n = 10, Probability of getting line busy (success) = 0.2. An experiment is nothing but a set of one or more repeated trials resulting in a particular outcome out of many outcomes. An unbiased coin is tossed 9 times. The binomial distribution X~Bin(n,p) is a probability distribution which results from the number of events in a sequence of n independent experiments with a binary / Boolean outcome: true or false, yes or no, event or no event, success or failure. The probability that atmost 3 families have television (X ≤ 3): ∴ P(X ≤ 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3), ∴ P(X ≤ 3) = 10C0 (0.8)0 (0.2)10 – 0 + 10C1 (0.8)1 (0.2)10 – 1 + 10C2 (0.8)2 (0.2)10 – 2 + 10C3 (0.8)3 (0.2)10 – 3, ∴ P(X ≤ 3) = 1 x 1 x (0.2)10 + 10 x (0.8) (0.2)9 + 45 x (0.8)2 (0.2)8 + 120 x (0.8)3 (0.2)7, Ans: The probability that 7 families have television is 0.2013 and the probability that atmost 3 families have television is 0.0008644. The probability of hitting the target atleast twice (X ≥ 2): ∴ P(X ≥ 2) = 1 – { 10C0 (0.2)0 (0.8)10 – 0 + 10C1 (0.2)1 (0.8)10 – 1}, ∴ P(X ≥ 2) = 1 – { 1 x 1 x (0.8)10 + 10 x (0.2) (0.8)9}, ∴ P(X ≥ 2) = 1 – (0.8 + 2) (0.8)9 = 1 – (2.8) (0.8)9, Ans: The probability of hitting the target atleast twice is 0.6242. The probability that none will recover (X = 0): The probability that all will recover (X = 6): The probability that halff of them will recover (X = 3): The probability that atleast half of them will recover (X ≥ 3): ∴ P(X ≥ 3) = P(X = 3) + P(X = 4) + P(x = 5) + P(X = 6), ∴ P(X ≥ 3) = 6C3 (0.7)3 (0.3)6 – 3 + 6C4 (0.7)4 (0.3)6 – 4 + 6C5 (0.7)5 (0.3)6 – 5 + 6C6 (0.7)6 (0.3)6 – 6, ∴ P(X ≥ 3) = 20 x (0.7)3 (0.3)3 + 10 x (0.7)4 (0.3)2 + 6 x (0.7)5 (0.3)1 + 1 x (0.7)6 (0.3)0, Ans: The probability that none will recover is 0.000729. b) atmost three correct answers? 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. In this case number of trials = n = 10, Probability of family having atelevision st is 80% = 80/100 = 0.8. In this article, we shall study to solve problems of probability based on the concept of the binomial distribution. An unbiased coin is tossed 5 times. An unbiased coin is tossed 8 times. In other words, the Bernoulli distribution is the binomial distribution that has a value of n=1.” The Bernoulli distribution is the set of the Bernoulli experiment. The probability that a person picked at random will support a constitutional amendment requiring an annual balanced budget is 0.8. The probability that atleast half of them will recover is 0.9294, Centres for disease control have determined that when a person is given a vaccine, the probability that the person will develop immunity to a virus is 0.8. In this case number of trials = n = 10, Probability of hitting target (success) = 0.8, Exactly two miss the target implies 8 bombs hit the target. We have only 2 possible incomes. atleast 6 supports the ammendment. In this case number of trials = n = 5, Probability of getting correct answer (success) = 1/4. Hence machine will be working if less than three components fail. 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. The binomial distribution is a discrete probability distribution that represents the probabilities of binomial random variables in a binomial experiment. The probability of getting atleast one head (X ≥ 1): ∴ P(X ≥ 1) = 1 – 1/256 = 255/256 = 0.9961, Ans: The probability of getting exactly 5 heads is 7/32 or 0.2188, The probability of getting a head a larger number of times than the tail is 93/256 or 0.3633, The probability of getting atleast one head is 255/256 or 0.9961. If eight people are given the vaccine, find the probability that a) none will develop immunity, b) exactly one will develop immunity, and c) all will develop immunity, In this case number of trials = n = 8, Probability taht person develops immunity (success) = 0.78. The probability that 7 families have television (X = 7): ∴ P(X = 7) = 120 x (0.8)7 (0.2)3 = 0.2013. In a town, 80% of all the families own a television set. The probability of getting atleast 1 correct answers (X ≥ 1): ∴ P(X ≥ 1) = 1 – 5C0 (1/4)0 (3/4)5 – 0 3, ∴ P(X ≥ 1) = 1- (243/1024) = 781/1024 = 0.7627, Ans: The probability of getting exactly 3 answers correct is 45/512 or 0.0879, the probability of getting atmost 3 correct answers is 63/64 or 0.9844, the probability of getting atleast 1 correct answer is 781/1024 or 0.7627, The probability of hitting a target in any shot is 0.2. The probability distribution of the random variable X is called a binomial distribution, and is given by the formula: \displaystyle {P} {\left ({X}\right)}= { {C}_ { {x}}^ { {n}}} {p}^ {x} {q}^ { { {n}- {x}}} P (X) = C xn The student is attempting to guess the answers. 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). In this article, we shall study to solve problems of probability based on the concept of the binomial distribution. Example – 01: An unbiased coin is tossed 5 times. Find the probability of getting a) three heads b) at least 4 heads. The number of successes X in n trials of a binomial experiment is called a binomial random variable. Every trial has a possible result, selected from S (for success), F (for failure), and each trial’s probability would be the same. A single success/failure test is also called a Bernoulli trial or Bernoulli experiment and a series of outcomes is called a Bernoulli process. The parameters …