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Binomial Distribution What the binomial distribution is How to recognise situations where the binomial distribution applies How to find probabilities for a given binomial distribution, by calculation and from tables When to use the binomial distribution Independent variables Pascal’s Triangle n (a+b) 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 nCr 5C0 1 5C1 5 5C2 10 5C3 10 5C4 5 5C5 1 10 ways to get to the 3rd position numbering each of the terms from 0 to 5. this can also be calculated by using nCr button on your calculator 5C2=10 Pascal’s Triangle n (a+b) 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 nCr n!÷(c!x(n-c)!) 5C0 5!÷(0!x5!) 1 5C1 5!÷(1!x4!) 5 5C2 5!÷(2!x3!) 10 5C3 5!÷(3!x2!) 10 5C4 5!÷(4!x1!) 5 5C5 5!÷(5!x0!) 1 A coin is tossed 7 times. Find the probability of getting exactly 3 heads. We could do Pascal's triangle or we could calculate: 7C3 x (P(H))7 The probability of getting a head is ½ n nCr r 7 7C 3 3 7 7 1 1 35 35 7 0.27 2 128 3 2 TASK Exercise A Page 61 Unequal Probabilities A dice is rolled 5 times What is the probability it will show 6 exactly 3 times? P(6’)=5/6 P(6)=1/6 5 5C 3 10 3 5 1 5 P(3 sixes in 5 rolls) 3 6 6 3 2 Task / Homework Exercise B Page 62 The Binomial distribution is all about success and failure. When to use the Binomial Distribution – – A fixed number ofX trials Only two outcomes – – (true, false; heads tails; girl,boy; six, not six …..) Each trial is independent IF the random variable X has Binomial distribution, then we write X ̴ B(n,p) Sometimes you have to use the Binomial Formula n x ( n x ) P( X x) p q , x where q 1 p Eggs are packed in boxes of 12. The probability that each egg is broken is 0.35 Find the probability in a random box of eggs: there are 4 broken eggs 12 P( X 4) 0.354 0.65(124) 495 0.354 0.658 4 0.235 to 3 significan t figures Task / homework Exercise C Page 65 Eggs are packed in boxes of 12. The probability that each egg is broken is 0.35 Find the probability in a random box of eggs: There are less than 3 broken eggs P( X 3) P( X 0) P( X 1) P( X 2) 12 12 12 0 (12) 1 (11) 0.35 0.65 0.35 0.65 0.352 0.65(10) 0 1 2 11 0.005688 12 0.351 0.6511 66 0.1225 0.01346 0.0151 USING TABLES of the Binomial distribution An easier way to add up binomial probabilities is to use the cumulative binomial tables Find the probability of getting 3 successes in 6 trials, when n=6 and p=0.3 n=6 x 0 1 2 3 4 5 6 P=0.3 P(X=x) 0.1176 0.4202 0.7443 0.9295 0.9891 0.9993 1.000 n=6 x 0 1 2 3 4 5 6 P=0.3 P(X=x) 0.1176 0.4202 0.7443 0.9295 0.9891 0.9993 1.000 http://assets.cambridge.org/97805216/05397/excerpt/9780521605397_excerpt.pdf The probability of getting 3 or fewer successes is found by adding: P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.1176 + 0.3026 + 0.3241 + 0.1852 = 0.9295 The probability of getting 3 or fewer successes is found by adding: P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.1176 + 0.3026 + 0.3241 + 0.1852 = 0.9295 This is a cumulative probability. Task / homework Exercise D page 67 Mean variance and standard deviation μ = Σx x P(X=x)=mean This is the description of how to get the mean of a discrete and random variable defined in previous chapter. The mean of a random variable whos distribution is B(n,p) is given as: μ =np Mean, variance & standard deviation σ²=Σx² x P(X=x) - μ² is the definition of variance, from the last chapter of a discrete random variable. The variance of a random variable whose distribution is B(n,p) σ²= np(1-p) σ= np(1 p) TASK / HOMEWORK Exercise E Mixed Questions Test Your self