University:
Stanford University
Course:
MATH 77Q | Probability and gambling
Academic year:

2021
Views:

21

Pages:

13
Author:

MathPaladin

Questions and Answers Sheet 4 Binomial Probability Question #1 Assume a binomial probability distribution has p=.60 and n=200 . a. what are the mean and standard deviation? b. Is this situation one in which binomial probabilities can be approximated by the normal probability distribution? Explain. c. what is the probability of 100 to 110 successes? Answer: bit a we are suppose to do first three sub parts Given, a binomial probability distribution has p=0.60 and n=200 a) Mean (𝜇) = 𝑛𝑝 𝜇 = 200 ⋅ 0.60 𝜇 = 120 Standard deviation (𝜎) = √𝑛𝑃(1 − 𝑃) 𝜎 = √200 ⋅ 0.60 ⋅ (1 − 0.60) 𝜎 = 6.9282 bit b b) Here 𝑛𝑃 = 200 ⋅ 0.60 nP=120>5 𝑛(1 − 𝑃) = 200 ⋅ (1 − 0.60) = 200 ⋅ 0.40 n(1-P)=80>5 both of which are more than 5 Hence normal distribution can be a good aproximation bit c c) Required probability is P(10013)=P(X>13.5) =1-P(X<13.5) = 1 − P( X − np 13.5 − (20  0.8)  ) npq 20  0.8  0.2 =1-P(z<-1.39754) =1-0.081125 =0.918875 Since the area corresponding to z-value -1.39754 is 0.081125 calculated by excel command NORM.S.DIST(-1.39754,TRUE). Thus the probability is 0.919. Question #4 1. A binomial distribution is the probability distribution of the type of probability experiment that has the three characteristic below: 2. The random variable X of a binomial probability experiment is .... 3. The mean \mu of a binomial distribution with n trials and the probability of success p= .... 4. The variance 𝜎 2 = ..... and the standard deviation 𝜎 = ....... Answer: Step 1 1. Characteristics of Binomial Distribution a) Number of Observation (n) is fixed b) The Probability of success (p) is equal for all the trials c) Every trial is independent of each other d) There are only two outcomes of the event ( success or failure) Step 2 2. The Random Variable X of a binomial probability experiment is number of successes in the 'n' trials 3. The mean μ of a binomial distribution with n trials and the probability of success 𝑝 = 𝑛 × 𝑝 Step 3 4. The Variance 𝜎 2 = 𝑛 × 𝑝 × (1 − 𝑝) = 𝑛 × 𝑝 × 𝑞 and the standard deviation = 𝑛 × 𝑝 × (1 − 𝑝) = √𝑛 × 𝑝 × 𝑞 Question #5 A binomial probability experiment is conducted with the given parameters. Compute the probability of x successes in the n independent trials of the experiment. n=10, p=0.6, x=7 P(7)=? Answer: Step 1 Given Information: Number of trials (n)=10 Probability of success (p) = 0.6 . To compute the probability of X=7 successes: Binomial formula: 𝑃(𝑋 = 𝑥) =𝑛 𝐶𝑥 𝑝 𝑥 (1 − 𝑝)𝑛−𝑥 Step 2 Required probability is obtained as follows: 10! 𝑃(𝑋 = 7) =10 𝐶7 (0.6)7 (1 − 0.6)10−7 = × 0.0279936 × (0.4)3 7! (10 − 7)! = 120 × 0.0279936 × 0.064 =0.214990848 ≈ 0.2150 Therefore, required probability is 0.2150 Question #6 If you play roulettes and bet on 'red' the probability that you win is 18/38 = .4737. People often repeat this between several times. We can consider each time we play a 'trial' and consider it a success when we win, so p = 18/38 or (.4737) and q = 20/38 or (.5263). Suppose that Caryl always places the same bet when she plays roulette, $5 on 'red'. Caryl might play just once, or might play several times. She has a profit (having won $5 more times than she lost $5) if she wins more than half of the games she plays. a) what is the probability that Caryl has a profit if she plays only one time? b) suppose Caryl has decided that she is going to play 13 times. to find the probability that has a profit, we can treat this like a binomial. Show the calculator input and result as your calculator to find the probability that she makes a profit when she plays 13 times. Answer: Step 1 Here, the probability of winning a game is 0.4737. if she win the game , then she will get $5. Step 2 a. Caryl plays only one time. Therefore, there is one chance either she will win or lose the game. If she win the game then she will get $5 without any additional amount. If she win more than one game, then she will get profit. In case, if she lose the first game for the first time will lose the $5 amount. Therefore, the probability of winning a game is 0.4737. Step 3 b. consider X is a random variable that represents the number of games played (if wins, she played more than half of the games). Step 4 Therefore, the binomial probability (X>6) is, P(X>6)=1-P(X \le 6) =1-0.5767 [=BINOMLIST (6,13,0.4737,TRUE)] =0.4233 Step 5 The probability that she has profit is 0.4233. Question #7 A binomial probability is given. Write the probability in words. Then, use a continuity correction to convert the binomial probability to a normal distribution probability. P(x=114) Write the probability in words. The probability of getting ? 114 successes. Which of the following is the normal probability statement that corresponds to the binomial probability statement? A) P(113.5114.5) E) P(x>113.5) Answer: Step 1 Given probability is P(x=114) The probability of getting exactly equal to 114 successes Step 2 P(X=114)=P(114-0.5 5) = 𝑃(𝑋 = 6) + 𝑃(𝑋 = 7) + 𝑃(𝑋 = 8) 8 8 8  ( )  0.86  (1 − 0.8)8−6 + ( )  0.87  (1 − 0.8)8−7 + ( )  (1 − 0.8)8−8 6 7 8 ⇒ 0.294 + 0.336 + 0.168 =0.797 Step 3 c) probability that the number that say they would feel secure is at most five is given by Applying Binomial theorem Probability of x success out of n trial is given by n P( X = x) = ( )  p x  (1 − p)n− x x ⇒ 𝑃(𝑋 ≤ 5) = 1 − 𝑃(𝑋 > 5) ⇒ 1 − 0.797 =0.203 Step 4 Answer a) probability that the number that say they would feel secure is exactly five is 0.1468 b) probability that the number that say they would feel secure is more than five is 0.797 c) probability that the number that say they would feel secure is at most five is 0.203 Question #13 Calculate the probability of exactly 3 successes in 4 tries. p=0.2 Answer: Step 1 Given: n=4, r=3, p=0.2 Consider that the random variable X follows a binomial distribution with parameters n and p. So, the binomial probability is, 𝑃(𝑋 = 𝑟) =𝑛 𝐶𝑟 × 𝑝𝑟 × (1 − 𝑝)𝑛−𝑟 Step 2 Here, r is the number of successes that results from the binomial experiment, n is the number of trials in the binomial experiment, and p is the probability of success on an individual trial. Thus, the probability of exactly 3 successes in 4 trials can be computed as: 𝑃(𝑋 = 3)4 𝐶3 × 0.23 (1 − 0.2)4−3 = 4 × 0.23 × 0.8 =0.0256 Step 3 Thus, the required probability is 0.0256. Question #14 Assume that a procedure yields a binomial distribution with a trial repeated n=5 times. Use some form of technology to find the cumulative probability distribution given the probability p=0.16 of success on a single trial. (Report answers accurate to 4 decimal places.) k P( X = k ) 0 1 2 3 4 5 Answer: Step 1 The binomial probability distribution is, n P( X = k ) = ( )( p)k (1 − p) n−k x In the formula, n denotes the number of trails, p denotes probability of success, and k denotes the number of success. Step 2 The probability of k=0 is, 5 P( X = 0) = ( )(0.16)0 (1 − 0.16)5−0 =0.4182 0 The probability of k = 1 is, 5 P( X = 1) = ( )(0.16)1 (1 − 0.16)5−1 1 =0.3983 The probability of k=2 is, 5 P( X = 2) = ( )(0.16)2 (1 − 0.16)5−2 2 =0.1517 The probability of k=3 is, 5 P( X = 3) = ( )(0.16)3 (1 − 0.16)5−3 3 =0.0289 The probability of k = 4 is, 5 P( X = 4) = ( )(0.16)4 (1 − 0.16)5−4 4 =0.0028 The probability of k = 5 is, 5 P( X = 5) = ( )(0.16)5 (1 − 0.16)5−5 5 =0.0001 Thus, the probability distribution given the probability p=0.16 of success on a single trial is, k P( X = k ) 0 0.4182 1 0.3983 2 0.1517 3 0.0289 4 0.0028 5 0.0001 Question #15 An agent sells life insurance policies to five equally aged, healthy people. According to recent date, the probability that a person lives in these conditions for 30 years or more is Then the probability that after 30 years exactly two people are still living is: Answer: Step 1 Given: Probability that a peon lives in these condition 30 years or more = 3/5 = 0.6 Sample size , n=5 equally aged people Let X denotes the number of people still living after 30 years X represents the Binomial random variable Step 2 3 5 . We need to compute a binomial distribution probability. The following information are provided: Population Probability of Success (p) = 0.6 Sample Size (n) = 5 Probability Event = exactly 2 people still living after 30 years Pr ( X  3) the binomial probability formula: n P( X ) = ( )  p X  (1 − p)n− X X This implies that 5 Pr ( X = 2) = ( )0.62  0.45−2 2 =0.2304 which means that the probability we are looking for is Pr(X=2)=0.2304 P (exactly 2 people still living after 30 years ) = 0.2304 Question #16 A binomial probability experiment is conducted with the given parameters. Compute the probability of x successes in the n independent trials of the experiment. n=5, p=.35, x=2 P(2) Answer: Step 1 Given: The experiment conducted is a binomial probability experiment. The information provided is: The number of experiments are n=5 . The probability of success is p=0.35 . Let x be a random variable following binomial distribution. It can be mathematically represented as: 𝑋 ∼ 𝐵𝑖𝑛(𝑛, 𝑝) ∼ 𝐵𝑖𝑛(5,0.35) The probability mass function of the random variable X is given as: 𝑃(𝑋 = 𝑥) =𝑛 𝐶𝑥 𝑝 𝑥 (1 − 𝑝)𝑛−𝑥 Step 2 The probability for 2 successes is given as: 𝑃(𝑋 = 2) =5 𝐶2 (0.35)2 (1 − 0.35)5−2 = 5! 2!(5−2)! × 0.352 × 0.653 = 10 × 0.1225 × 0.274625 =0.336415 ∼ 0.3364 Therefore, the required probability is P(2)=0.3364 . Question #17 In a large population, about 65% of families take a vacation during the summer. A researcher takes a random sample of 16 families and surveys whether they take a summer vacation. Use the binomial distribution to compute the probability that exactly 8 of the families take a summer vacation. Identify the following information required to find the probability of families that take a vacation during the summer. Provide your answer below: n= trials x= successes p= probability of summer vacation Answer: Step 1 The binomial probability distribution is, n P( X = x) = ( )( p) x (1 − p) n− x x In the formula, n denotes the number of trails, p denotes probability of success, and x denotes the number of success. The random variable X is defined as the number of families take a summer vacation which follows normal distribution with sample size 16 and probability of success 0.65. The probability that exactly 8 of the families take a summer vacation have to find. That is, n=16, x=8, p=0.65 . Step 2 The probability that exactly 8 of the families take a summer vacation is, 16 P( X = 8) = ( )(0.65)8 (1 − 0.65)16−8 8 = 12,870 × 0.03186 × 0.000225 =0.0923 Thus, the probability that exactly 8 of the families take a summer vacation is 0.0923. Question #18 Suppose that X has a hypergeometric distribution with N=500, n=50, and K=100 . Find P(X=10) by using the hypergeometric distribution and approximate binomial distribution. Is the binomial approximation reasonable? Answer: Step 1 Hypergeometric Distribution: N: Number of items in the population K: Number of items in the population classified as success n: Number of items in the sample x: Number of items in the sample classified as success The hypergeometric probability distribution is: h(x,N,n,k)=\frac{(10010)(500-10050-10)}{(50050)} = (10010)(40040) (50050) Since the calculation is time consuming, use online hypergeometric calculator or Excel to find the probability. Using Excel function, “=HYPGEOM.DIST(10,50,100,500,FALSE)”, the probability is 0.1474. Step 2 Binomial distribution: n: Sample size p: Probability of success x: Number of items in the sample classified as success Binomial probability distribution is, 𝐵(𝑥; 𝑛, 𝑝) = (𝑛𝑥)𝑝 𝑥 (1 − 𝑝)𝑛−𝑥 where p=K/N . Given: N=500, n=50, K=100 and x=10 . That is, p=100/500 =0.2 . Thus, the Binomial distribution is to be obtained for, 𝐵(10; 50,0.2) = (5010)0.210 (1 − 0.2)50−10 = (5010)0.210 (0.8)40 =0.1398 The hypergeometric probability 0.1474 is pretty close to the Binomial approximation 0.1398. Thus, the Binomial approximation is reasonable. Question #19 If you toss a fair coin 11 times, what is the probability of getting all heads? Express the probability as a simplified fraction. 1 A) 4096 1 B) C) D) 2048 1 2 1 1024 Answer: Step 1 A fair coin is tossed. There are two possible outcomes for this event as appearing a head or tail. Coin is tossed 11 times and probability of occurrence of head is (1/2) and for tail is (1/2). Probability of getting all heads can be calculated using binomial probability. A random variable x is said to follow binomial distribution if its probability density function is given as: 𝑃(𝑋 = 𝑥) =𝑛 𝐶𝑥 𝑝 𝑥 (1 − 𝑝)𝑛−𝑥 Step 2 X is event that getting all heads in 11 tosses. For tossing 11 coins, probability of getting all heads can be calculated as: 𝑃(𝑋 = 11) =11 𝐶11 𝑝11 (1 − 𝑝)11−11 1 11 =( ) = 2 1 2,048 Therefore, the correct answer is ''Option B, 1/2048''. Question #20 To obtain: The probability that the claim will be accepted. Given info: The sample size is n=100 and the drug manufacturer claims that a drug cures a rare skin disease is 65% of the time. Answer: Calculation: The drug cures a rare skin disease p is 0.65. The drug not cures a rare skin disease q is 0.35. The requirements to check whether the normal distribution can be used for approximate the distribution of x, the binomial random variable are 𝑛𝑝 ≥ 5 and 𝑛𝑞 ≥ 5 . That is, Condition 1: 𝑛𝑝 ≥ 5 Condition 2: 𝑛𝑞 ≥ 5 Condition 1: 𝑛𝑝 ≥ 5 Substitute 100 for n and 0.65 for p in the np. 𝑛𝑝 = 100 × 0.65 =65 >5 Thus, the given requirement 𝑛𝑝(= 65) ≥ 5 is satisfied. Condition 2: 𝑛𝑞 ≥ 5 Substitute 100 for n and 0.35 for q in the nq. 𝑛𝑞 = 100 × 0.35 =35 >5 Thus, the given requirement 𝑛𝑞(= 35) ≥ 5 is satisfied. Since the requirements that 𝑛𝑝 ≥ 5 and 𝑛𝑞 ≥ 5 both are satisfied. Thus, the normal distribution can be used to approximate the binomial distribution. That is, the probability from a binomial probability distribution can be approximated by using normal distribution with the parameters are, 𝜇 − 𝑛𝑝 and 𝜎 = √𝑛𝑝𝑞. The value of mean and standard deviation is, 𝜇 = 100 × 0.65 =65 𝜎 = √100 × 0.65 × 0.35 = √22.75 =4.7697 Here, the claim will be accepted represent that the binomial random variable x is at least 70. Continuity correction: If the binomial probability represents ''at least c'' then the normal probability is P(x>c-0.5) for any number c. By using continuity correction, the value 0.5 is subtracted from the value of 70. That is, c=70 𝑃(𝑥 ≥ 70) = 𝑃(𝑥 > 70 − 0.5) = 𝑃(𝑥 > 69.5) Here, P(x>69.5) represents the area to the right of 69.5. The formula to convert the x value into z score is, 𝑥−𝜇 𝑧= 𝜎 Substitute 69.5 for x, 65 for 𝜇 and 4.7697 for 𝜎 𝑧= = 69.5−65 4.7697 4.5 4.7697 =0.94 The probability that the claim will be accepted is obtained by finding area to the right of 0.94. But Table 4, the Standard normal distribution is used to find the areas to the left. Use Table 4: Standard normal distribution to find the area to the left of 0.94. Procedure: For z at 0.94, - Locate 0.9 in the left column of the Table 4. - Obtain the value in the corresponding row below 0.04. That is, 𝑃(𝑧 < 0.94) = 0.8264 The area to the right of 0.94 is, 𝑃(𝑧 > 0.94) = 1 − 𝑃(𝑧 < 0.94) =1-0.8264 =0.1736 Thus, the probability that the claim will be accepted is 0.1736.