Binomial probability #10 - Questions and Answers

Question 1 In addition to his social studies class, Edward must complete a five-question multiple-choice exam. There were four possible answers for each question, but only one of them was right. Assuming Edward guesses on each of the five inquiries, what is the likelihood that he will respond correctly to all five questions, exactly two questions, or at least two questions? Answer 1 Step 1 Given: Total number of questions 𝑛 = 5 n=5. Probability of success 𝑝 = 1 4 p= 4 1 ​ (since each question has only 1 correct choice out of 4). We use the binomial distribution here. Step 2 (a) The probability that he will answer all five questions correctly: 𝑃 ( 𝑋 = 5 ) = ( 5 5 ) ( 1 4 ) 5 ( 1 − 1 4 ) 5 − 5 P(X=5)=( 5 5 ​ )( 4 1 ​ ) 5 (1− 4 1 ​ ) 5−5 = 1 ⋅ ( 1 4 ) 5 ⋅ 1 = 0.000977 =1⋅( 4 1 ​ ) 5 ⋅1=0.000977 (b) The probability that he will answer exactly 2 questions correctly: 𝑃 ( 𝑋 = 2 ) = ( 5 2 ) ( 1 4 ) 2 ( 1 − 1 4 ) 5 − 2 P(X=2)=( 2 5 ​ )( 4 1 ​ ) 2 (1− 4 1 ​ ) 5−2 𝑃 ( 𝑋 = 2 ) = 10 ⋅ ( 1 4 ) 2 ⋅ ( 3 4 ) 3 P(X=2)=10⋅( 4 1 ​ ) 2 ⋅( 4 3 ​ ) 3 𝑃 ( 𝑋 = 2 ) = 0.26367 P(X=2)=0.26367 (c) The probability that he will answer at least 2 questions correctly: 𝑃 ( 𝑋 ≥ 2 ) = 1 − 𝑃 ( 𝑋 < 2 ) = 1 − 𝑃 ( 𝑋 = 0 ) − 𝑃 ( 𝑋 = 1 ) P(X≥2)=1−P(X<2)=1−P(X=0)−P(X=1) 𝑃 ( 𝑋 = 0 ) = ( 5 0 ) ( 1 4 ) 0 ( 3 4 ) 5 = 0.2373 P(X=0)=( 0 5 ​ )( 4 1 ​ ) 0 ( 4 3 ​ ) 5 =0.2373 𝑃 ( 𝑋 = 1 ) = ( 5 1 ) ( 1 4 ) 1 ( 3 4 ) 4 = 0.3955 P(X=1)=( 1 5 ​ )( 4 1 ​ ) 1 ( 4 3 ​ ) 4 =0.3955 𝑃 ( 𝑋 ≥ 2 ) = 1 − [ 0.2373 + 0.3955 ] = 1 − 0.6328 = 0.3672 P(X≥2)=1−[0.2373+0.3955]=1−0.6328=0.3672 Answer 2 Step 1 (a) The probability that Edward will answer all five questions correctly: 𝑃 ( Six correct answers ) = ( 6 6 ) ( 1 4 ) 6 ( 1 − 1 4 ) 0 ≈ 0.0002 P(Six correct answers)=( 6 6 ​ )( 4 1 ​ ) 6 (1− 4 1 ​ ) 0 ≈0.0002 Step 2 (b) The probability that Edward will answer exactly two questions correctly: 𝑃 ( Two correct answers ) = ( 6 2 ) ( 1 4 ) 2 ( 3 4 ) 4 ≈ 0.2966 P(Two correct answers)=( 2 6 ​ )( 4 1 ​ ) 2 ( 4 3 ​ ) 4 ≈0.2966 Step 3 The probability that Edward will answer at least two questions correctly: 𝑃 ( 𝑋 ≥ 2 ) = 1 − 𝑃 ( 𝑋 < 2 ) = 1 − 𝑃 ( 𝑋 = 0 ) − 𝑃 ( 𝑋 = 1 ) P(X≥2)=1−P(X<2)=1−P(X=0)−P(X=1) 𝑃 ( 𝑋 = 0 ) = ( 6 0 ) ( 1 4 ) 0 ( 3 4 ) 6 = 0.1780 P(X=0)=( 0 6 ​ )( 4 1 ​ ) 0 ( 4 3 ​ ) 6 =0.1780 𝑃 ( 𝑋 = 1 ) = ( 6 1 ) ( 1 4 ) 1 ( 3 4 ) 5 = 0.3560 P(X=1)=( 1 6 ​ )( 4 1 ​ ) 1 ( 4 3 ​ ) 5 =0.3560 𝑃 ( 𝑋 ≥ 2 ) = 1 − [ 0.1780 + 0.3560 ] = 0.4660 P(X≥2)=1−[0.1780+0.3560]=0.4660 Thus, the probability that Edward will answer at least two questions correctly is 0.4660. Question 2 What does exclamation on point mean? Answer 1 In mathematics, the symbol represents the factorial operation. Answer 2 It’s a factorial denoted by the exclamation mark (!). Question 3 Alexis rolls an 8-sided die 24 times. What is the probability a “five” is rolled at least 4 times, but no more than 18 times? Answer Explanation: The binomial probability relation is: ∑ 𝑘 = 0 𝑛 𝐶 𝑛 𝑘 𝑝 𝑘 ( 1 − 𝑝 ) 𝑛 − 𝑘 = 1 k=0 ∑ n ​ C n k ​ p k (1−p) n−k =1 Here 𝑛 = 24 n=24, 𝑝 = 1 8 p= 8 1 ​ , and we’re looking at 4 ≤ 𝑘 ≤ 18 4≤k≤18. Using cumulative probability: Find 𝑃 ( 𝑘 ≤ 3 ) P(k≤3) and 𝑃 ( 𝑘 ≥ 19 ) P(k≥19). Subtract these from 1 to find 4 ≤ 𝑘 ≤ 18 4≤k≤18. The cumulative probability for up to 18 successes is approximately 0.6476. The cumulative probability for up to 3 successes is 0.3524. Thus, the probability is roughly: 0.6476 − 0.3524 = 0.3524 0.6476−0.3524=0.3524 Question 4 What is the variance of a binomial distribution for which 𝑛 = 75 n=75 and 𝑝 = 0.20 p=0.20? Answer 1 The formula for the variance of a binomial distribution is: Variance = 𝑛 ⋅ 𝑝 ⋅ 𝑞 where   𝑞 = 1 − 𝑝 Variance=n⋅p⋅qwhereq=1−p Here 𝑛 = 75 n=75, 𝑝 = 0.20 p=0.20, 𝑞 = 1 − 0.20 = 0.80 q=1−0.20=0.80: Variance = 75 ⋅ 0.20 ⋅ 0.80 = 12 Variance=75⋅0.20⋅0.80=12 Answer 2 Note that: If 𝑋 ∼ 𝐵 ( 𝑛 , 𝑝 ) X∼B(n,p), then Variance of 𝑋 X is 𝑛 𝑝 ( 1 − 𝑝 ) np(1−p). Variance = 75 ⋅ 0.20 ⋅ 0.80 = 12 Variance=75⋅0.20⋅0.80=12 Answer: 12 Question 5 How do you find the probability of at least one success when 𝑛 n independent Bernoulli trials are carried out with probability of success 𝑝 p? Answer 1 The probability is: 𝑃 ( 𝑋 = 𝑘 ) = ( 𝑛 𝑘 ) 𝑝 𝑘 𝑞 𝑛 − 𝑘 , where   𝑞 = 1 − 𝑝 P(X=k)=( k n ​ )p k q n−k ,whereq=1−p Thus, the probability of at least one success is: 𝑃 ( 𝑋 ≥ 1 ) = 1 − 𝑃 ( 𝑋 = 0 ) P(X≥1)=1−P(X=0) 𝑃 ( 𝑋 = 0 ) = ( 1 − 𝑝 ) 𝑛 P(X=0)=(1−p) n 𝑃 ( 𝑋 ≥ 1 ) = 1 − ( 1 − 𝑝 ) 𝑛 P(X≥1)=1−(1−p) n Answer 2 To calculate the probability of getting at least one success, use the complement rule: 𝑃 ( 𝑋 ≥ 1 ) = 1 − 𝑃 ( 𝑋 = 0 ) P(X≥1)=1−P(X=0) The probability that there are no successes is: 𝑃 ( 𝑋 = 0 ) = ( 1 − 𝑝 ) 𝑛 P(X=0)=(1−p) n Thus: 𝑃 ( 𝑋 ≥ 1 ) = 1 − ( 1 − 𝑝 ) 𝑛 P(X≥1)=1−(1−p) n