Binomial Probability Distribution SUB-TOPICS: Definition and basic concept. Definition and basic concept. The binomial distribution arises from a repeated random experiment which has two possible outcomes: a) That an event will occur and b) That an event will not occur In the experiment of tossing a fair coin repeatedly, there are two possible outcomes namely; a head or a tail. Each repetition of tossing a coin is called a trial. The two possible outcomes of the experiment are usually called success and failure. If the event that a head comes up is a success, then the event that a tail shows up is a failure and vice versa. Probability of success is usually denoted p and the probability of a failure by q; and p+q=1 or q=1- p. If we denote the probability that in n trials, we have r successes and therefore (n-r) failures by pr(x=r), then p(x=r)= nCrp r q n-r .(Note nCr = ) The parameters of the binomial distribution are n and p. Properties of the Binomial Distribution If mean is denoted by μ and standard deviation by σ then (i) Mean μ = np (ii) Standard deviation σ = √npq (iii) Variance σ2 = npq Examples: 1. Find the probability that when a fair coin is tossed three times, a head shows up twice. Solution: Note that in a single toss of a coin the probability of a head showing up is ½ and therefore of a head not showing up is 1- 1 /2 = ½. So if p is the probability of success (i.e. of a head showing up), then p=1/2 and therefore q=1/2. The coin is tossed three times, hence n=3. Since the required is a head showing up twice (2 times) then r=2. Applying the formula pr(x=r)= nCrp r q n-r Pr(2)= 3C2(1/2)2 (1/2)1 = 3/8 2. Find the probability that in 10 throws of a fair six faced die, a square number shows up three times. Solution: Let p be the prob that a square number shows up. Since there are only two square numbers from 1 to 6 (i.e. 1 and 4), then p=2/6=1/3, Therefore q=4/6=2/3 (ie 1- 2 /6); The die is thrown ten times implying n=10; and r=3 (the number of times the square numbers are expected to turn up) Applying the formula Pr(3)=10C3(1/3)3 (2/3)7 = 5120/19683 = 0.26 3. In an examination, 60% of the candidates passed. Calculate the prob that a random sample of 10 candidates contains at most two failures. Solution: Let p be the probability that a candidate fails the exam then p=4/10; therefore q, the probability that a candidate passes the exam is 6/10 (ie 1- 4 /10) At most 2 failures interprets to no failure,1 failure or 2 failures.That is r=0, r=1, r=2. Pr(0)=10C0(2/5)0 (3/5)10 …………..(i) Pr(1)=10C1(2/5)1 (3/5)9…………….(ii) Pr(2)=10C2(2/5)2 (3/5)8……………..(iii) Therefore the required probability is (i)+(ii)+(iii) = 0.167 ASSIGNMENT 1 1. Find the probability that when a fair six-faced die is tossed eight times, a prime number appears exactly four times. 2. When a biased coin is tossed, the probability that a head shows up is two thirds. Find the probability that when the coin is tossed 7 times a head will appear exactly 5 times. 3. The probability that a diabetic patient survives when injected with a newly discovered drug is 0.75. Find the probability that exactly 8 out of 10 diabetic patients survive on being injected with the drug. 4. If 20% of bolts produced by a machine are defective, determine the probability, correct to two decimal places, that out of 8 bolts chosen at random: (i) One is defective (ii) At most two are defective (iii) Exactly four are effective.

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