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12.1 Discrete Probability Distributions (Binomial Distribution) 1 At the end of the lesson, students will be able to (c)Understand the Binomial distribution B(n,p) (d) find the mean and variance of Binomial distribution. 2 BERNOULLI TRIAL Bernoulli trial is a trial that has 2 possible outcomes or 3 The Bernoulli trial is repeated n times and each trial is independent. The probability of a success stays the same for each trial. The number of successes (X) are recorded as a random variable. Intro Bernoulli 4 Probability distribution for X is a binomial distribution Denoted by X ~ B( n, p ) where n = the number of independent trials p = the probability of success X~B(n,p) read as ‘X is binomially distributed where n is the number of trials and p is the probability of success’ 5 The binomial probability distribution is given by n x n- x P(X = x) = Cxp q x = 0, 1, 2, 3, … , n Where x = the number of success n = the number of trials p = the probability of success q=1–p Binomial Example 6 EXAMPLE 1 A fair coin is tossed three times. Find the probability of getting (a) no head, (b) exactly one head, (c) exactly two heads, (d) exactly three heads. 7 SOLUTION If X represents the number of heads obtained, then X~B(3, 0.5) P(X=x) = n x Cxp (1 - p) (a) P(X = 0) = (b) P(X = 1) = 3 n- x 0 3- 0 = 0.125 1 3- 1 = 0.375 C0 (0.5) (0.5) 3 C1 (0.5) (0.5) 8 (c) P(X = 2) = 3 2 3- 2 = 0.375 3 3- 3 = 0.125 C2 (0.5) (0.5) 3 (d) P(X = 3) = C3 (0.5) (0.5) 9 EXAMPLE 2 In a box, 20% of bulbs are damaged. If a random sample of 10 bulbs is taken, find the probability that (a) exactly 3 bulbs are damaged. (b) all bulbs are good. (c) not more three bulbs are damaged. 10 SOLUTION If X represents the number of damaged bulbs, then X~B(10, 0.2) (a) P(X = 3) = 10 10 3 7 C3 (0.2) (0.8) = 0.2013 0 10 (b) P(X = 0) = C0 (0.2) (0.8) = 0.1074 11 (c) P(X ≤ 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 10 C0 (0.2)0 (0.8)10 + + 10 C2 (0.2)2 (0.8)8 + 10 10 C1 (0.2)1 (0.8)9 C3 (0.2)3 (0.8)7 = 0.1074 + 0.2684 + 0.3020 + 0.2013 = 0.8791 12 EXAMPLE 3 The probability that students watch a certain TV program is 0.4. If eight students are selected at random, find the probability that (a) 5 students watch the TV program. (b) less than 4 students watch the TV program. (c) at least one student watches the TV program. 13 SOLUTION If X represents the number of students watch a certain TV program, then X~B(8, 0.4) and 8 x 8- x P(X = x) = Cx (0.4) (0.6) (a) P(X = 5) = 8 5 8- 5 5 3 C5 (0.4) (0.6) 8 = C5 (0.4) (0.6) = 0.1239 14 (b) P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) 8 0 8 8 1 = C0 (0.4) (0.6) + C1 (0.4) (0.6) 8 2 6 8 3 7 + C2 (0.4) (0.6) + C3 (0.4) (0.6) 5 = 0.01679 + 0.08957 + 0.20901 + 0.27869 = 0.5941 15 (c) P(X ≥ 1) = 1 – P(X < 1) = 1 – P(X = 0) = 1- 8 0 C0 (0.4) (0.6) 8 = 0.9832 16 MEAN AND VARIANCE OF BINOMIAL DISTRIBUTION If X ~ B(n, p), then (i) mean, μ = E(X) = np, (ii) variances, σ2 = Var(X) = npq *Notes: q = 1 – p 17 EXAMPLE 4 A fair die is rolled 300 times. Let X represents the number of times the number ‘6’ is observed. Find the mean and standard deviation. 18 SOLUTION If X represents the number of times the number 6 is observed, then æ X ~ B çç300, çè ö 1÷ ÷ ø 6÷ æ1 ö ÷ = 50 Mean, μ = np = (300) ç çç ÷ è6 ÷ ø 19 æ1 öæ5 ö ÷ = 41.67 Variance, σ2 = npq = (300) ççç ÷ç ÷ç ÷ è 6 ÷ç øè 6 ÷ ø Standard Deviation, σ = = Var(X) 41.67 = 6.45 20 EXAMPLE 5 A study revealed that two of five families have a video recorder. (a) If there are 4 families, find the probability that two or more families have video recorders. (b) If there are 900 families, calculate the mean and standard deviation of the numbers of families who have video recorders 21 SOLUTION If X represents the number of families who have video recorders, then X~B(4, 0.4) and P(X = x) = 4 Cx (0.4)x (0.6)44 x 2 2 4 3 1 (a) P(X ≥ 2) = C2 (0.4) (0.6) + C3 (0.4) (0.6) 4 4 + C4 (0.4) (0.6) 0 = 0.5248 22 (b) X~B(900, 0.4) μ = np = (900)(0.4) = 360 σ= npq = (900)(0.4)(0.6) = 14.70 23 BINOMIAL PROBABILITY TABLES To find probability for the value of p from 0.001 to 0.5 n P(X ³ r) = å n Crp (1 - p) r n- r X= r P(X ≥ r) is ‘Probability of X is larger and Equals to r. 24 Binomial Cumulative Distribution F(x) P(X r) 1 P(X r) 1 P(X r 1) 25 EXAMPLE 6 Let X be a random such that X~B(5, 0.3). By using the binomial table, find (a) P(X ≥ 3) (b) P(X > 3) (c) P(X ≤ 3) (d) P(X < 3) (e) P(X = 3) 26 SOLUTION (Turn to page 5 from Statistical Tables) n p 0.10 0.15 0.20 0.25 0.30 0.35 r 5 0 1 2 3 4 1.000 1.000 1.000 0.4095 0.5563 0.6723 0.0815 0.1648 0.2627 0.0086 0.0266 0.0579 0.0005 0.0022 0.0067 (a) P(X ≥ 3) = 0.1631 1.000 1.000 1.000 0.7627 0.8319 0.8840 0.3672 0.4718 0.5716 0.1035 0.1631 0.2352 0.0156 0.0308 0.0540 Check Table 27 (b) P(X > 3) = P(X ≥ 4) = 0.0308 (c) P(X ≤ 3) = 1 – P(X ≥ 4) = 0.9692 (d) P(X < 3) = 1 – P(X ≥ 3) = 0.8369 (e) P(X = 3) = P(X ≥ 3) – P(X ≥ 4) = 0.1631 – 0.0308 = 0.1323 Check Table 28 TO FIND PROBABILITY FOR THE VALUE OF P > 0.5 (i) P[ X r / X ii P[ X r / X iii P[ X r / X B(n, p)] P[ X n r / X B(n,1 p)] B(n, p )] P[ X n r / X B (n,1 p )] B(n, p)] P[ X n r / X B (n,1 p )] 29 EXAMPLE 7 Let X be a random variable such that X~B(10, 0.8). By using the binomial table, find (a) P(X ≥ 4) (b) P(X > 4) (c) P(X ≤ 4) (d) P(X < 4) 30 SOLUTION (a) P(X ≥ 4) P(X ≤ 6) = 1 - P(X ≥ 7) = 1 - 0.0009 = 0.9991 (b) P (X > 4) = P(X ≥ 5) P (X ≤ 5) = 1 - P(X ≥ 6) = 1 - 0.0064 = 0.9936 X ~ B(10,0.8) X ~ B(10,0.2) X ~ B(10,0.8) X ~ B(10,0.2) 31 (c) P(X ≤ 4) P(X ≥ 6) = 0.0064 X ~ B(10,0.8) X ~ B(10,0.2) (d) P(X < 4) = P (X ≤ 3 ) P(X ≥ 7) X ~ B(10,0.8) X ~ B(10,0.2) = 0.0009 32 EXAMPLE 8 For a random variable X with a binomial distribution B(10, 0.45), find a when (a) P(X ≥ a) = 0.4956 (b) P(X < a) = 0.89801 (c) P(X = a) = 0.23836 33 SOLUTION (a) P(X ≥ a) = 0.4956 from the binomial probability tables, P(X ≥ 5) = 0.4956 Thus, a = 5 34 (b) P(X < a) = 0.89801 1 – P(X ≥ a) = 0.89801 P(X ≥ a) = 1 – 0.89801 = 0.1020 From the binomial probability tables, P(X ≥ 7) = 0.1020 Thus, a = 7 35 (c) P(X = a) = 0.23836 P(X ≥ a) – P(X ≥ a + 1) = 0.23836 From the binomial probability tables, P(X ≥ 4) – P(X ≥ 5) = 0.23836 Thus, a = 4 36 CONCLUSION The binomial probability distribution is given n x n- x by P(X = x) = Cxp q x = 0, 1, 2, 3, … , n To find probability for the value of p > 0.5 (i) P[ X r / X B(n, p)] P[ X n r / X B(n,1 P) ii P[ X r / X iii P[ X r / X B(n, p)] P[ X n r / X B(n,1 P) B(n, p)] P[ X n r / X B(n,1 P) 37

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