Binomial probability #5 - Questions and Answers

Question 1 To match: For a continuity adjustment, the binomial probability statement and its equivalent normal distribution probability statement (a)–(d). Answer The reason this is the right response is because the binomial random variable 𝑥 x is greater than 109. The value of 109 is increased by the value of 0.5 by the use of continuity correction. That is, 𝑥 = 109 x=109 𝑃 ( 𝑥 > 109 ) = 𝑃 ( 𝑥 > 109 + 0.5 ) = 𝑃 ( 𝑥 > 109.5 ) P(x>109)=P(x>109+0.5)=P(x>109.5) Thus, the binomial probability statement corresponding to the normal distribution probability statement is 𝑃 ( 𝑥 > 109.5 ) P(x>109.5). The wrong answers are (b), (c), and (d) because the binomial probability statement, which corresponds to the normal distribution probability statement, is (b), (c), and (d), respectively, 𝑃 ( 𝑥 > 109.5 ) P(x>109.5). Therefore, the correct option is (a). Question 2 Assume that random guesses are made for eight multiple choice questions on an SAT test, so that there are 𝑛 = 8 n=8 trials, each with probability of success (correct) given by 𝑝 = 0.20 p=0.20. Find the indicated probability for the number of correct answers. Find the probability that the number 𝑥 x of correct answers is fewer than 3. Answer Step 1 Binomial Probability: The formula for binomial probability is as follows: 𝑃 ( 𝑥 ) = 𝐶 𝑛 𝑥 𝑝 𝑥 ( 1 − 𝑝 ) 𝑛 − 𝑥 P(x)=C n x ​ p x (1−p) n−x Step 2 Probability that the number 𝑥 x of correct answers is fewer than 3: Here 𝑥 x follows binomial distribution. Here, 𝑛 = 8 n=8; 𝑝 = 0.20 p=0.20. The probability that the number 𝑥 x of correct answers is fewer than 3 is obtained as 0.797, as shown below: 𝑃 ( fewer than 3 ) = 𝑃 ( 0 ) + 𝑃 ( 1 ) + 𝑃 ( 2 ) P(fewer than 3)=P(0)+P(1)+P(2) = [ 𝐶 8 0 ( 0.2 ) 0 ( 1 − 0.2 ) 8 ] + [ 𝐶 8 1 ( 0.2 ) 1 ( 1 − 0.2 ) 7 ] + [ 𝐶 8 2 ( 0.2 ) 2 ( 1 − 0.2 ) 6 ] =[C 8 0 ​ (0.2) 0 (1−0.2) 8 ]+[C 8 1 ​ (0.2) 1 (1−0.2) 7 ]+[C 8 2 ​ (0.2) 2 (1−0.2) 6 ] = 0.168 + 0.335 + 0.294 =0.168+0.335+0.294 = 0.797 =0.797 Question 3 I have a large jar of red and green M&M’s with 55% red M&M’s. I plan to pick 7 M&M’s. I want to find the probability that I get 2 red M&M’s out of 7. a) How many different ways can you get 2 red M&M’s out of 7? Use the formula for the binomial coefficient to find that number. Answer Step 1 The binomial coefficient is computed by 𝐶 𝑛 𝑘 C n k ​ , which gives the number of ways to select 𝑘 k objects from 𝑛 n objects. Here it is asked to find the number of ways to draw 2 red candy out of 7, so substitute 𝑛 = 7 n=7 and 𝑘 = 2 k=2 in the formula of binomial coefficient. Number of ways = 𝐶 7 2 C 7 2 ​ : 𝐶 7 2 = 7 ! 2 ! ( 7 − 2 ) ! = 7 ⋅ 6 2 ⋅ 1 = 21 C 7 2 ​ = 2!(7−2)! 7! ​ = 2⋅1 7⋅6 ​ =21 Step 2 The probability of binomial random variable is computed by formula: 𝑝 ( 𝑘 ) = 𝐶 𝑛 𝑘 𝑝 𝑘 ( 1 − 𝑝 ) 𝑛 − 𝑘 , p(k)=C n k ​ p k (1−p) n−k , where 𝑛 n is the total number of trials, 𝑘 k is the number of successes, and 𝑝 p is the probability of success in each trial. In each trial, the probability of getting red candy is 55% or 0.55 and there are total 7 trials out of which 2 should be successful (drawn red). So substitute 𝑛 = 7 , 𝑘 = 2 , 𝑝 = 0.55 n=7,k=2,p=0.55: 𝑃 ( 2 ) = 𝐶 7 2 ( 0.55 ) 2 ( 1 − 0.55 ) 7 − 2 P(2)=C 7 2 ​ (0.55) 2 (1−0.55) 7−2 = 21 ⋅ 0.3025 ⋅ 0.4 5 5 =21⋅0.3025⋅0.45 5 ≈ 0.1172 ≈0.1172 Step 3 For the case of 3 red candies out of 7, the total number of trials are 𝑛 = 7 n=7, number of successes are 𝑘 = 3 k=3, and probability of success is 𝑝 = 0.55 p=0.55. To find the probability of 3 successes, follow the steps mentioned below on a calculator: Select and add 𝑏 𝑖 𝑛 𝑜 𝑚 𝑝 𝑑 𝑓 ( 7 , 0.55 , 3 ) binompdf(7,0.55,3). Then add the parameters in this sequence 7 , 0.55 , 3 7,0.55,3. Press Enter. This will give the probability of getting 3 red candies out of 7, which is equal to 0.2388. Question 4 The variance for the binomial probability distribution is: − ( 𝑥 ) = 𝑃 ( 1 − 𝑃 ) −(x)=P(1−P) − ( 𝑥 ) = 𝑛 𝑃 −(x)=nP − ( 𝑥 ) = 𝑛 ( 1 − 𝑃 ) −(x)=n(1−P) − ( 𝑥 ) = 𝑛 𝑃 ( 1 − 𝑃 ) −(x)=nP(1−P) Answer Step 1: Given information Given that the distribution of binomial probabilities: 𝑥 x follows Binomial ( 𝑛 , 𝑃 n,P) Step 2: Answer is given by: 𝑥 ∼ Binomial ( 𝑛 , 𝑃 ) x∼Binomial(n,P) Then: Mean = 𝑛 𝑃 Mean=nP Variance = 𝑛 𝑃 ( 1 − 𝑃 ) Variance=nP(1−P) Therefore, Option (4) is correct. Question 5 This exercise looks at how well the Poisson distribution approximates the binomial distribution. Assume that 𝑛 = 363 n=363 and 𝑝 = 0.003 p=0.003 for a binomial distribution. Find the exact (actual) probability (using the binomial probability formula) of one success in 363 trials. (Report answer accurate to 6 decimal places.) 𝑃 ( 𝑥 = 1 ) P(x=1). Find the approximate probability of one success in 363 trials using Poisson distribution. 𝑃 ( 𝑥 = 1 ) P(x=1). Find the relative difference between the two values. (Use all of the decimal places shown in your calculator from the previous two answers). Answer Step 1 Given 𝑛 = 363 n=363 and 𝑝 = 0.003 p=0.003. The probability mass function of Binomial distribution is: 𝑃 ( 𝑋 = 𝑥 ) = 𝐶 𝑛 𝑥 𝑝 𝑥 ( 1 − 𝑝 ) 𝑛 − 𝑥 P(X=x)=C n x ​ p x (1−p) n−x For 𝑥 = 1 x=1: 𝑃 ( 𝑋 = 1 ) = 363 ! ( 363 − 1 ) ! 1 ! ( 0.003 ) 1 ( 0.997 ) 362 P(X=1)= (363−1)!1! 363! ​ (0.003) 1 (0.997) 362 𝑃 ( 𝑋 = 1 ) = 363 ⋅ 0.003 ⋅ 0.33701351 P(X=1)=363⋅0.003⋅0.33701351 𝑃 ( 𝑋 = 1 ) = 0.367008 P(X=1)=0.367008 Step 2 Using Poisson distribution: Here, 𝜆 = 𝑛 ⋅ 𝑝 = 363 ⋅ 0.003 = 1.089 λ=n⋅p=363⋅0.003=1.089. The Poisson formula: 𝑃 ( 𝑋 = 𝑥 ) = 𝑒 − 𝜆 𝜆 𝑥 𝑥 ! P(X=x)= x! e −λ λ x ​ For 𝑋 = 1 X=1: 𝑃 ( 𝑋 = 1 ) = 𝑒 − 1.089 ⋅ ( 1.089 ) 1 1 ! P(X=1)=e −1.089 ⋅ 1! (1.089) 1 ​ 𝑃 ( 𝑋 = 1 ) = 𝑒 − 1.089 ⋅ 1.089 P(X=1)=e −1.089 ⋅1.089