Questions and Answers #7 Binomial Probability

Questions and Answers Sheet 7 Binomial Probability Question #1 It is believed that 58% of Americans like to eat at restaurants every week. You interview 6 Americans and ask if they like to eat at restaurants every week. The random variable x represents the number of Americans that like to eat at restaurants every week. Find the mean of the binomial distribution. 𝜇 =? Find the standard deviation of the binomial distribution. 𝜎 = ? What is the probability that at least 1 of the Americans surveyed likes to eat at restaurants every week? Provide an answer to 4 decimal places. Answer: Step 1 Given that: Number of observations = 𝑛 = 6 Probability of success = 𝑝 = 58% = 0.58 So, X follows binomial distribution with 𝑛 = 6 and 𝑝 = 0.58 𝑚𝑒𝑎𝑛 = 𝜇 = 𝑛𝑝 = 6(0.58) = 3.48 standard deviation =𝜎 = √𝑛𝑝(1 − 𝑝) = √6(0.58)(1 − 0.58) = √1.4616 = 1.20896650078 ≈ 1.2090 Step 2 At least 1 means 1 or more. We will use the complement property of probability for this 𝑃(𝑋 ≥ 1) = 1 − 𝑃(𝑋 = 0) = 1 − (6𝐶0)(0.58)0 (1 − 0.58)6 = 1 − 0.005489031744 = 0.994510968256 ≈ 0.9945 Answer: a) 3.48 b) 1.2090 c) 0.9945 Question #2 Study the binomial distribution table. Notice that the probability of success on a single trial p ranges from 0.01 to 0.95. Some binomial distribution tables stop at 0.50 because of the symmetry in the table. Let's look for that symmetry. Consider the section of the table for which 𝑛 = 5 . Look at the numbers in the columns headed by 𝑝 = 0.30 and 𝑝 = 0.70 . Do you detect any similarities? Consider the following probabilities for a binomial experiment with five trials. (a) Compare P(3 successes), where 𝑝 = 0.30 , with P(2 successes) , where 𝑝 = 0.70 . P(3 successes) , where 𝑝 = 0.30 , is larger. P(3 successes), where 𝑝 = 0.30 , is smaller. They are the same. (b) Compare P(3 or more successes), where 𝑝 = 0.30 , with P(2 or fewer successes), where 𝑝 = 0.70 . P(3 or more successes), where 𝑝 = 0.30 , is larger.P(3 or more successes), where 𝑝 = 0.30 , is smaller. They are the same. (c) Find the value of P(4 successes), where 𝑝 = 0.30 . (Round your answer to three decimal places.) P(4 successes) = For what value of r is P(r successes) the same using 𝑝 = 0.70? r= Answer: Step 1 a) Given that 𝑛 = 5 Consider P(3 successes), if 𝑝 = 0.30  5 P(3 successes) =   0.33 (1 − 0.3)5−3 = 0.1323  3 Consider P(2 or fewer successes), where 𝑝 = 0.7 P(2 or fewer successes) = P(0 success) + P(1 success) + P(2 success) 5  5 5 =   0.70 (1 − 0.7)5−0 +   0.71 (1 − 0.7)5−1 +   0.7 2 (1 − 0.7)5−2 0 1  2 = 0.1631 Here, the value of P(3 successes) is less than P(2 or fewer successes). P(3 successes), where 𝑝 = 0.30 , is smaller than P(2 or fewer successes), where p = 0.70 . Step 2 b) Given that 𝑛 = 5 Consider P(3 or more successes), if 𝑝 = 0.30 P(3 or more successes)  5 5  5 =   0.33 (1 − 0.3)5−3 +   0.34 (1 − 0.3)5−4 +   0.35 (1 − 0.3)5−5  3  4  5 = 0.1631 Consider P(2 or fewer successes), where 𝑝 = 0.7 P(2 or fewer successes) = P(0 success) + P(1 success) + P(2 success) 5 5 5 =   0.70 (1 − 0.7)5−0 +   0.71 (1 − 0.7)5−1 +   0.7 2 (1 − 0.7)5−2 = 0.1631 0 1  2 Here, the value of P(3 or more successes) is same as P(2 or fewer successes). P(3 or more successes), where p = 0.30 , is same as P(2 or fewer successes), where 𝑝 = 0.70 . Step 3 c) P(4 successes), where p = 0.30 . 5 P(4 successes) = ( )0.34 (1 − 0.3)5−4 = 0.028 4 Question #3 How do you use the binomial probability formula to find the probability of x successes given the probability p of success on a single trial for 𝑛 = 5, 𝑥 = 2, 𝑝 = 0.25 ? Answer: Step 1 Let p(x) denote the reqd. prod. Then, 𝑝(𝑥) =𝑛 𝐶𝑥 𝑝 𝑥 𝑞 𝑛−𝑥 𝑥 = 0,1,2, ⋯ , 𝑏; &, 𝑞 = (1 − 𝑝) In our case, 𝑛 = 5, 𝑥 = 2, 𝑝 = 0.25, & so, 𝑞 = 1 − 0.25 = 0.75 ∴ 𝑝(2) =5 𝐶2 (0.25)2 (0.75)5−2 5×4 1 2 3 3 = ( ) ( ) 1×2 4 4 1 27 = 10 × × 16 64 5 × 27 = 8 × 64 135 𝑝(2) = 512 Question #4 How do you use the binomial probability formula to find the probability x of success for a given probability p of success in one trial for 𝑛 = 6, 𝑥 = 4, 𝑝 = 0.75 ? Answer: Step 1 𝑃(𝑋 = 4), 𝑛 = 6, 𝑝 = 0.75 𝑋 ∼ Binomial(𝑛 = 6, 𝑝 = 0.75) 𝑃(𝑋 = 4) = ((𝑛), (𝑥))𝑝 𝑥 (1 − 𝑝)𝑛−𝑥 = ((6), (4))(0.75)4 (1 − 0.75)6−4 = 15 × (0.75)4 (0.25)2 = 0.296631 ≈ 0.296631 ≈ 0.297 𝑃(𝑋 = 4) = 0.297 Question #5 He probability of rain on any given day is 2 3 . What is the probability of at most one day of rain during the next 3 days? Answer: Step 1 At most one day of rain during next three days means rain on not more than one day during next three days. Let R denote rain on the day and N denote no rain on the day. Further let XXX denote R or N on next three days. Ten we are looking for a pattern i.e RNN, NRN, NNR or NNN 2 1 Now as the probability of rain i.e. R is , probability of N is 3 3 and hence 𝑃(𝑅𝑁𝑁) = 𝑃(𝑁𝑅𝑁) = 𝑃(𝑁𝑁𝑅) 2 1 1 2 = × × = 3 3 3 27 and 1 1 1 1 𝑃(𝑁𝑁𝑁) = × × = 3 3 3 27 Hence, the probability of at most one day of rain during next three day is 2 1 7 3× + = 27 27 27 or about 26% Question #6 Assume that a procedure yields a binomial distribution. What is the probability of x successes if 𝑛 = 64, 𝑥 = 3 , and 𝑝 = 0.04 ? Answer: Step 1 𝑃(𝑋 = 3), 𝑛 = 64, 𝑝 = 0.04 𝑋 ∼ Binomial(𝑛 = 64, 𝑝 = 0.04) P( X = 3) = ((n), ( x)) p x (1 − p ) n − x 64 = ( )(0.04)3 (1 − 0.04)64−3 3 = 41664  (0.04)3 (0.96) 61 = 0.221048  0.221048  0.221 𝑃(𝑋 = 3) = 0.221 Question #7 How do you find the mean, variance and standard deviation of the binomial distribution with 𝑛 = 316and 𝑝 = 0.72 ? Answer: The mean is = 227.52 The variance is = 63.71 The standard deviation is = 7.98 Explanation: The binomial distribution with 𝑛 = 316and𝑝 = 0.72 The mean is 𝜇𝑥 = 𝑛 ⋅ 𝑝 = 316 ⋅ 0.72 = 227.52 The variance is = 𝜎𝑥2 = 𝑛 ⋅ 𝑝 ⋅ (1 − 𝑝) = 316 ⋅ 0.72 ⋅ 0.28 = 63.71 The standard deviation is = 𝜎𝑥 = √𝑛 ⋅ 𝑝 ⋅ (1 − 𝑝) = √63.7056 = 7.98 Question #8 What is the mean of the binomial distribution for which 𝑛 = 20and𝑝 = 0.2 ? Answer: For a binomial distribution 𝑋 − 𝐵(𝑛, 𝑝) the mean is 𝐸(𝑋) = 𝑛𝑝,and its variance𝑉𝑎𝑟(𝑋) = 𝑛𝑝(1 − 𝑝) in this case we have: 𝑋 − 𝐵(20,0,2) for good measure the variance will be 𝑉𝑎𝑟(𝑋) = 20 × 0.2 × 0.8 = 3.2 Question #9 1 Consider Bernoulli trials with success probability 𝑝 = . What is the probability that the first success 4 occurs in the fourth trial? Answer: We can approach this by first seeing that with the probability of success being is 3 4 1 4 , the probability of failure . Next we need to remember that each trial is independent. We have a specific set of outcomes that we're solving for: (Fail, Fail, Fail, Success) and with the related probabilities of each: 3 3 3 1 ( , , , ) 4 4 4 4 we can multiply each probability to find an overall probability of the first success on the 4th attempt: 3 3 3 1 27 4 4 4 4 256 ( × × × )= ≅ 10.55% Question #10 For binomial distribution X, with 𝑛 = 7and𝑝 = 0.6 , what is 𝑃(𝑋 > 3) ? Answer: 𝑃𝑟{𝑋 > 3} = 𝑓(4) + 𝑓(5) + 𝑓(6) + 𝑓(7) = (0.6)4 [35 × (0.4)3 + 21 × 0.6 × (0.4)2 + 7 × (0.6)2 × 0.4 + (0.6)3 ] = 0.710208 Explanation: 𝑓(𝑥) = 7𝐶𝑥 (0.6) 𝑥 (0.4)7−𝑥 , 𝑥 = 0,1, … 7 X follows Binomial Distribution with parameters 𝑝 = 0.6and𝑛 = 7 . 5 7𝐶4 = 7𝐶3 = 7 × 6 × × 2 × 3 = 7 × 5 = 35 6 1 7𝐶5 = 7𝐶2 = 7 × × 2 = 21,7𝐶6 = 7𝐶1 = 7 and 7𝐶7 = 1. 1 Question #11 What is the variance of a binomial distribution for which 𝑛 = 75and𝑝 = 0.20 ? Answer: 1 4 5 5 75 × ( ) × ( ) Explanation: The formula for the variance of a binomial distribution is 𝑛 × 𝑝 × 𝑞 where 𝑞 = 1 − 𝑝 . Here 𝑛 = 75, 𝑝 = 1 4 5 5 0.20 or ( ) and 𝑞 = 0.80or ( ) and hence the answer. Question #12 Of the pupils in a school, 30% travel to school by bus. From a sample of 10 pupils chosen at random, what is the probability that more than 8 travel by bus? Answer: Step 1 We have a binomial probability. I like to start these kinds of questions with this relation: ∑𝑛𝑘=0 𝐶𝑛, (𝑝)𝑘 (1 − 𝑝)𝑛−𝑘 = 1We have 𝑛 = 10, 𝑝 = 0.3 adn are looking at 8 < 𝑘 ≤ 10 𝐶10, 9 (0.3)9 (0.7)1 + 𝐶10, 10(0.3)10 (0.7)0 ≈ 0.00014 Question #13 65% of men consider themselves knowledgeable soccer fans. If 14 men are randomly selected, what is the probability that exactly 4 of them will consider themselves knowledgeable fans? Answer: Step 1 𝑃(𝑋 = 4), 𝑛 = 14, 𝑝 = 0.65 𝑋 ∼ Binomial(𝑛 = 14, 𝑝 = 0.65) 𝑃(𝑋 = 4) = ((𝑛), (𝑥))𝑝 𝑥 (1 − 𝑝)𝑛−𝑥 = ((14), (4))(0.65)4 (1 − 0.65)14−4 = 1001 × (0.65)4 (0.35)10 = 0.004929 ≈ 0.004929 ≈ 0.005 𝑃(𝑋 = 4) = 0.005 Question #14 What is P(x) for 𝑛 = 4, 𝑥 = 1, 𝑝 = 0.6 ? Answer: Step 1 The probability can be calculated as: 𝑃(𝑥) = ((𝑛), (𝑥)) × 𝑝 𝑥 (1 − 𝑝)𝑛−𝑥 Here it is: 𝑃(1) = ((4), (1)) × (0.6)1 (0.4)3 𝑃(1) = 4 × 0.6 × 0.43 = 2.4 × 0.064 = 0.1536 Question #15 What is P(x) for 𝑛 = 6, 𝑥 = 3, 𝑞 = 0.7 ? Answer: Step 1 𝑃(𝑥) =𝑛 𝐶𝑥 𝑝 𝑥 𝑞 𝑛−𝑥 𝑥 = 0, 1, 2, 3, ⋯ , 𝑛 and 𝑝 + 𝑞 = 1 We have, 𝑛 = 6, 𝑥 = 3 and 𝑞 = 0.7 ⇒ 𝑝 = 1 − 𝑞 = 0.3 ∴ 𝑃(3) =6 𝐶3 (0.3)3 (0.7){6−3) = (20)(0.3)3 (0.7)3 = 20(0.21)3 ∴ 𝑃(3) = 20(0.009261) = 0.18522 Question #16 What is P(x) for 𝑛 = 3, 𝑥 = 0, 𝑝 = 0.8 ? Answer: Step 1 For the Binomial Distribution, 𝑃(𝑥) =𝑛 𝐶𝑥 𝑝 𝑥 𝑞 𝑛−𝑥 𝑥 = 0, 1, 2, ⋯ , 𝑛; 𝑞 = (1 − 𝑝) In our Example, 𝑛 = 3, 𝑥 = 0, &, 𝑝 = 0.8 ⇒ 𝑞 = 1 − 𝑝 = 0.2 ∴ 𝑃(0) =3 𝐶0 (0.8)0 (0.2)3−0 = (1)(1)(0.008) = 0.008 Question #17 What is P(x) for 𝑛 = 4, 𝑥 = 2, 𝑝 = 0.4 ? Answer: Step 1 𝑃(𝑋 = 2), 𝑛 = 4𝑝 = 0.4 𝑋 ∼ Binomial(𝑛 = 4, 𝑝 = 0.4) 𝑃(𝑋 = 2) = ((𝑛), (𝑥))𝑝 𝑥 (1 − 𝑝)𝑛−𝑥 = ((4), (2))(0.4)2 (1 − 0.4)4−2 = 6 × (0.4)2 (0.6)2 = 0.3456 = 0.345600 ≈ 0.346 𝑃(𝑋 = 2) = 0.346 Question #18 What is P(x) for 𝑛 = 6, 𝑥 = 3, 𝑝 = 0.3 ? Answer: Step 1 𝑃(𝑋 = 3), 𝑛 = 6𝑝 = 1 − 𝑞 = 0.7 𝑋 ∼ Binomial(𝑛 = 6, 𝑝 = 0.7) 𝑃(𝑋 = 3) = ((𝑛), (𝑥))𝑝 𝑥 (1 − 𝑝)𝑛−𝑥 = ((6), (3))(0.7)3 (1 − 0.7){6−3) = 20 × (0.7)3 (0.3)3 = 0.18522 = 0.185220 ≈ 0.185 𝑃(𝑋 = 3) = 0.185 Question #19 What is P(x) for 𝑛 = 3, 𝑥 = 1, 𝑝 = 0.9 ? Answer: Step 1 For a Binomial Random Variable X with parameters n and p, then 𝑃(𝑋 = 𝑥) = 𝑝(𝑥) =𝑛 𝐶𝑥 𝑝𝑛 𝑞 𝑛−𝑥 𝑥 = 0, 1, 2, ⋯ , 𝑛 Here, 𝑛 = 3, 𝑝 = 0.9, 𝑥 = 1 Also, 𝑞 = 1 − 𝑝 = 1 − 0.9 = 0.1 ∴ 𝑝(1) =3 𝐶1 (0.9)1 (0.1)3−1 = 3(0.9)(0.1)2 = 3(0.9)(0.01) ⇒ 𝑝(1) = 0.027 Question #20 If x is a binomial random variable, what is the probability of x for 𝑛 = 4, 𝑥 = 1, 𝑝 = 0.3 ? Answer: Step 1 𝑃(𝑋 = 1), 𝑛 = 4𝑝 = 0.3 𝑋 ∼ Binomial(𝑛 = 4, 𝑝 = 0.3) 𝑃(𝑋 = 1) = ((𝑛), (𝑥))𝑝 𝑥 (1 − 𝑝)𝑛−𝑥 = ((4), (1))(0.3)1 (1 − 0.3)4−1 = 4 × (0.3)1 (0.7)3 = 0.4116 = 0.411600 ≈ 0.412 𝑃(𝑋 = 1) = 0.412