Chpt 5 5-4 The Binomial Distribution 1 /36
Chpt 5-4 Chpt 5 Homework p262 Applying the Concepts Exercises p263 1-11, 14-18, 23, 24, 26 2 /36
Objective Chpt 5 Find the exact probability for x successes in n trials of a binomial experiment. 3 /36
Binomial When there can be only two outcomes for an event, or when we can reduce the outcomes to two possibilities, we have a binomial experiment or Bernoulli trial. (Not the same Bernoulli as in Bernoulli s Principle) Many types of probability problems have only two possible outcomes or can be reduced to two outcomes. Essentially the two outcomes are: success or failure Is you is, or is you ain t H or T M or F Acceptable or Unacceptable or < 4 /36
Binomial Experiment A binomial experiment has 4 conditions 1. There is a finite number of repeated trials. 2. The results of each trial is only success or failure. Success is not necessarily good. Success simply means the result expected. Failure is anything other than the chosen result. 3. The trials must be independent. 4. The probability for success must remain constant for every trial. 5 /36
Binomial Probability To keep notation clear, we use the following conventions. P(S) = probability of success = p P(F) = probability of failure = q P(S) = 1 - P(F) or p = 1 q P(F) = 1 - P(S) or q = 1 - p X = number of successes in n trials (x = 0, 1, 2,... n) P(X=x) is the probability of exactly x number of successes out of n trials. 6 /36
Binomial Probability For example, suppose we roll a die four times. Success is rolling a 6. There are only two possible outcomes; 6, not 6. With 4 rolls, there are 2 4 = 16 possible results. Let X = not 6. 6, 6, 6, 6 6, 6, 6, X 6, 6, X, 6 6, X, 6, 6 X, 6, 6, 6 6, 6, X, X 6, X, 6, X 6, X, X, 6 X, 6, X, 6 X, X, 6, 6 X, 6, 6, X 6, X, X, X X, 6, X, X X, X, 6, X X, X, X, 6 X, X, X, X What would be the probability of rolling exactly one 6? p(6&x&x&x) = 1/6 5/6 5/6 5/6 = 125/1296 Not so fast bucko There are 4 ways to roll a single 6, all having the exact same probability. X, X, X, 6 or X, X, 6, X or X, 6, X, X or 6, X, X, X P(X = 1) = 125/1296 + 125/1296 + 125/1296 + 125/1296 = 500/1296 = 125/324 7 /36
Binomial Probability This gets complicated quickly, thus we move to make life easier. P (X = x ) = n! (n x )! x! p x q n x P(X=x) is the probability of exactly X number of successes out of n trials. 8 /36
Binomial Probability P(X) is the probability of exactly X number of successes out of n trials. P (X = x ) = n! (n x )! x! p x q n x To understand the formula let us look at the components. The number of ways we can get X successes (along with n X failures) out of n trials is simply the Combination of n results taken X at a time, ncx (order is not important). n C x = n! (n X )! X! The first component of the formula is the number of ways exactly x successes can occur in n trials. 9 /36
Binomial Probability P (X = x ) = n! (n x )! x! p x q n x Remember: P(a and b) = P(a) P(b) Probability of success = p, probability of exactly x successes = p x Probability of failure = q, probability of n x failures = q n-x The probability of each individual sequence of x successes AND (n x) failures is the product of the probabilities. In this case p x q n x. 10/36
Binomial Probability P (X = x ) = n! (n x )! x! p x q n x The number of ways we can get x successes (and n x failures) out of n trials is then multiplied by the probability of the x successes and n - x failures. Remember n = total number of trials x = number of successes p = probability of success n-x = number of failures q = probability of failure (1-p) P (X = x ) = n! (n x )! x! p x q n x 11/36
Binomial Probability Back to the probability for rolling exactly one 6 in 4 rolls of a die. P (X = x ) = n! (n x )! x! p x q n x P (X = 1) = 4! (4 1)!1! 1 6 1 5 6 3 = 4 1 6 125 216 = 500 1296 12/36
Example Remember this? For a binomial p must remain constant so let s pretend p(boy) =.51 3rd child 1st child p(b).51 p(g).49 2nd child p(b\b).51 p(g\b).49 p(b\g).51 p(g\g).49 p(b\bb).51 p(g\bb).49 p(b\bg).51 p(g\bg).49 p(b\gb).51 p(g\gb).49 p(b\gg).51 p(g\gg).49 p(bbb) =.51 3 =.132651 p(bbg) =.51 2.49 =.127449 p(bgb) =.51 2.49 =.127449 p(bgg) =.51.49 2 =.122451 p(gbb) =.51 2.49 =.127449 p(gbg) =.51.49 2 =.122451 p(ggb) =.51.49 2 =.122451 p(ggg) =.49 3 =.117649 P (bbb ) = C 3 0.51 3.49 0.1327 P (2b ) = C 3 2.51 2.49 1.3823 P (1b ) = C 3 1.51 1.49 2.3674 P (0b ) = C.51 0.49 3.1176 3 0 x P(x) 0 0.1176 1 0.3674 2 0.3823 3 0.1327 13/36
Example In a statistics class, students miss class 17% of class days. If the class has 12 students, find the probability of no absences on a given day, then find the probability of 3 absences on a given day. Only two possible results: absent, not absent. Success is absent, trials are independent, only 12 trials, and P(S) =.17 for all trials. This qualifies as a binomial experiment. P (X = x ) = n! (n x )! x! p x q n x 14/36
Example P (X = x ) = n! (n x )! x! p x q n x miss class 17%, 12 students, p(3absent), p(3absent)? n = 12, and p = P(absent) =.17, q = P(not absent) =.83. 0 absent: x = 0 P (0) = 12! (12 0)!0!.170.83 12 = 1 1.1069.107 3 absent: x = 3 P (3) = 12! (12 3)!3!.173.83 9 = 220.004913.18694.20206 15/36
Table s Rather than compute the actual probabilities using a calculator (there was a time calculators were not available) the probabilities can be looked up in a table. Beginning on page 756 (Table B) you can find a binomial distribution table. Simply find the value of n in the table, then find x, and finally find the correct p, and voila you have the probability. I tell you this so you will know tables are available should you ever be without your calculator. But we will use the calculator, so I will not go into using the tables. 16/36
TI-84 Easier to use the calculator. 2nd VARS (Distr) Alpha A or scroll down to A Binompdf(n,p,x) (Binomial probability density function) For the statistics class miss class 17%, 12 students, p(3absent), p(3absent)? Binompdf(12,.17, 0) =.1068900077 binompdf(12,.17, 3) =.2020562443 17/36
Binomial Distribution Binom(n,p) A probability distribution with only two outcomes is simply a Binomial Distribution. The outcomes of a binomial experiment and the corresponding probabilities of these outcomes make up a Binomial Distribution. x P(x) 0 p(x=0) A Binomial Distribution is defined by the number of attempts (n) and the probability of success (p). 1 p(x=1) 2 p(x=2) n p(x=n) 18/36
Example For the example of rolling a die 4 times. P (0) = 4! (4 0)!0! 1 6 0 5 6 4 = 1 1 625 1296 = 625 1296 P (X = x ) = n! (n x )! x! p x q n x P (1) P (2) P (3) P (4) = = = = 4! (4 1)!1! 1 6 4! (4 2)! 2! 1 6 4! (4 3)!3! 1 6 1 3 4! (4 4)! 4! 1 6 2 3 5 6 4 5 6 1 5 6 2 0 5 6 = 4 1 6 125 216 = 500 1296 = 6 1 36 25 36 = 150 1296 = 4 1 216 5 6 = 20 1296 = 1 1 1296 1= 1 1296 x P(x) 0 0.4823 1 0.3858 2 0.1157 3 0.0154 4 0.0008 19/36
Display The binomial distribution 4 Rolls of a Die x P(x) 0.5 0 0.4823 0.375 1 0.3858 2 0.1157 P(X) 0.25 3 0.0154 4 0.0008 0.125 0 0 1 2 3 4 Number of 6 rolled 20/36
Tree We can show the binomial of rolling 6 in 4 attempts using a tree diagram. 1st Roll p(6)=1/6 p(~6)=5/6 2nd Roll p(6)=1/6 p(~6)=5/6 p(6)=1/6 p(~6)=5/6 3rd Roll p(6)=1/6 p(~6)=5/6 p(6)=1/6 p(~6)=5/6 p(6)=1/6 p(~6)=5/6 p(6)=1/6 p(~6)=5/6 4th Roll p(6)=1/6 p(~6)=5/6 p(6)=1/6 p(~6)=5/6 p(6)=1/6 p(6)=1/6 p(~6)=5/6 p(~6)=5/6 p(6)=1/6 p(~6)=5/6 p(6)=1/6 p(~6)=5/6 p(6)=1/6 p(~6)=5/6 p(6)=1/6 p(~6)=5/6 p(6&6&6&6)=(1/6) 4.0008 p(6&6&6&~6)=(1/6) 3 (5/6).0039 p(6&6&~6&6)=(1/6) 3 (5/6).0039 p(6&6&~6&~6)=(1/6) 2 (5/6) 2.0193 p(6&~6&6&6)=(1/6) 3 (5/6).0039 p(6&6&~6&~6)=(1/6) 2 (5/6) 2.0193 p(6&6&~6&~6)=(1/6) 2 (5/6) 2.0193 p(6&~6&~6&~6)=(1/6)(5/6) 3.0965 p(~6&6&6&6)=(1/6) 3 (5/6).0039 p(~6&6&6&~6)=(1/6) 2 (5/6) 2.0193 p(~6&6&~6&6)=(1/6) 2 (5/6) 2.0193 p(~6&6&~6&~6)=(1/6)(5/6) 3.0965 p(~6&~6&6&6)=(1/6) 2 (5/6) 2.0193 p(~6&~6&6&~6)=(1/6)(5/6) 3.0965 p(~6&~6&~6&6)=(1/6)(5/6) 3.0965 p(~6&~6&~6&~6)=(5/6) 4.4823 21/36
Compare Results Comparing our tree results to our table. p(6&6&6&6)=(1/6) 4.0008 p(6&6&6&~6)=(1/6) 3 (5/6).0039 p(6&6&~6&6)=(1/6) 3 (5/6).0039 p(6&6&~6&~6)=(1/6) 2 (5/6) 2.0193 p(6&~6&6&6)=(1/6) 3 (5/6).0039 p(6&6&~6&~6)=(1/6) 2 (5/6) 2.0193 p(6&6&~6&~6)=(1/6) 2 (5/6) 2.0193 p(6&~6&~6&~6)=(1/6)(5/6) 3.0965 p(~6&6&6&6)=(1/6) 3 (5/6).0039 p(~6&6&6&~6)=(1/6) 2 (5/6) 2.0193 p(~6&6&~6&6)=(1/6) 2 (5/6) 2.0193 p(~6&6&~6&~6)=(1/6)(5/6) 3.0965 p(~6&~6&6&6)=(1/6) 2 (5/6) 2.0193 p(~6&~6&6&~6)=(1/6)(5/6) 3.0965 p(~6&~6&~6&6)=(1/6)(5/6) 3.0965 p(~6&~6&~6&~6)=(5/6) 4.4823 x P(x) 04 0.0008 0.4823 13 0.3858 0.0154 2 0.1157 31 0.0154 0.3858 40 0.0008 0.4823 p(exactly 4 6)=.0008 p(exactly 3 6)=4(.0039) p(exactly 2 6)=6(.0193) p(exactly 1 6)=4(.0965) p(exactly 0 6)=.4823 22/36
Example If a student randomly guesses at five (a-e) multiple-choice questions, find the probability that the student gets exactly three correct. Each question has five possible choices. We have a finite number of trials, n = 5 There are only two outcomes, correct, not correct The probability of correct remains constant at.2 The measures are independent (we assume). P (X = x ) = n! (n x )! x! p x q n x P (X = 3) = 5! (5 3)!3!.20 ( ) 3.80 ( ) 2 = 10.008.64 =.0512 23/36
Example Now lets make the problem a little more interesting. What is the probability the student will get at least 3 correct. This changes the problem. At least 3 correct means 3 correct or 4 correct or 5 correct. x P(x) 0 0.32768 1 0.4096 P (X = x ) = n! (n x )! x! p x q n x 2 0.2048 P (X 3) = P (X = 3) + P (X = 4) + P (X = 5) 3 0.0512 4 0.0064 P (X 3) =.0512 +.0064 +.00032 =.05792 5 0.00032 24/36
Mean and Variance The binomial distribution is a distribution of values (all possible number of successes). Thus, as any distribution, the binomial distribution has a mean and standard deviation. μ= np This is intuitive. If the probability of success is p and we have n trials, the likely number of successes would be np. σ 2 = npq (q = 1 - p) The algebraic derivation is more trouble than provides benefit, but we can easily show that the formulae work. 25/36
Example For the 4 rolls of a die the mean and variance can be calculated as with any probability distribution. x P(x) xp(x) (x-μ) (x-μ) 2 (x-μ) 2 p(x) 0 0.485 1 0.386 0 0.386-0.67 0.33 0.4489 0.1089 0.2177 0.042 µ = np = 4 1 6 = 2 3 2 0.116 0.232 1.33 1.7689 0.2052 3 0.015 0.045 2.33 5.4289 0.0814 σ 2 = npq = 4 1 6 5 6 = 5 9 4 0.0008 0.0032 3.33 11.0889 0.0089 μ=0.67 σ 2 =0.5552 26/36
Example In all the statistics classes, 17% of the students miss class each day. If there are 200 students taking statistics, find the mean and standard deviation for the number of daily absences. We have a finite number of trials, n = 200 There are only two outcomes, miss, do not miss µ = np = 200.17 ( ) = 34 σ 2 = npq = 200.17 ( )(.83) = 28.22 The probability of missing remains constant σ = npq = 28.22 5.3122 The measures are independent (we assume). There will be an average of 34 absences each day with a standard deviation of just over 5 absences. 27/36
TI-84 To find the probability of exactly x successes in n trials of a binomial experiment we use... 2nd - Vars (Distr) - A:binompdf(n,p,x) For our example of rolling a die 4 times and getting no (0) 1s binompdf(4,.1667, 0) =.4821759306 Getting one (1) exactly one time binompdf(4,.1667, 1) =.3858333 28/36
TI-84 To find the probability of at most x successes in n trials. 2nd - Vars (Distr) - B:binomcdf(n,p,x) Remember, the calculator will only calculate the probability of x successes or fewer. To calculate a probability of x or more you must find the complement and subtract from 1. To calculate a probability of 3 or more (X 3) you must find the probability of 2 or fewer and subtract from 1. (1 - binomcdf(n,p, 2)) To calculate a probability of more than 3 ( X > 3)you must find the probability of 3 or fewer and subtract from 1. (1 - binomcdf(n,p, 3)) 29/36
Example If a student randomly guesses at five (a-e) multiple-choice questions, find the probability that the student gets exactly three correct. Each question has five possible choices. What is the probability the student will guess at least 3 correct. x P(x) P (X 3) = P (X = 3) + P (X = 4) + P (X = 5) P (X 3) =.0512 +.0064 +.00032 =.05792 0 0.32768 1 0.4096 2 0.2048 The TI-84 finds the probability of at most x successes in n trials. 2nd - Vars (Distr) - B:binomcdf(n,p,x) For our example of at least 3 correct in 5 guesses 3 0.0512 4 0.0064 5 0.00032 binomcdf(5,.2, 2) =.94208 1 -.94208 =.05792 30/36
Example x P(x) 0 0.32768 1 0.4096 P (X 3) = P (X = 3) + P (X = 4) + P (X = 5) = 1 - binomcdf(5,.2, 2) = 1 -.94208 =.05792 2 0.2048 3 0.0512 4 0.0064 P (X > 3) = P (X = 4) + P (X = 5) =.0064 +.00032 =.00672 = 1 - binomcdf(5,.2, 3) = 1 -.99328 =.00672 5 0.00032 P (X 3) = P (X = 0) + P (X = 1) + P (X = 2) + p (x = 3) =.3277 +.4096 +.2048+.0512 =.9933 = binomcdf(5,.2, 3) =.9933 P (X < 3) = P (X = 0) + P (X = 1) + P (X = 2) =.3277 +.4096 +.2048 =.9421 = binomcdf(5,.2, 2) =.9421 31/36
The Geometric Model Geom(p) A Geometric probability model tells us the probability for a random variable that counts the number of Bernoulli trials until the first success. Geometric models are completely specified by one parameter, p, the probability of success, and are denoted Geom(p). 32/36
Geometric Model Geom(p) Geometric probability model for Bernoulli trials: Geom(p) n p = probability of success n q = 1 p = probability of failure n X = number of trials until the first success occurs P(X = x) = q x-1 p E (x ) = µ = 1 p σ = q p 2 33/36
TI-84 To find the geometric probability that the first success is on the xth trial 2nd - Vars (Distr) - E: geompdf(p,x) For our example of rolling a die and getting the first 6 on the 5th attempt geompdf(1/6, 5) =.0804 To find the geometric probability that the first success is on or before, the xth trial 2nd - Vars (Distr) - E: geomcdf(p,x) For our example of rolling a die and getting the first 6 on or before the 5th attempt geomcdf(1/6, 5) =.5981 34/36
Example Let us stipulate that your good friend Bart Simpson is a 60% student. That is, he averages 60% on tests. His teacher gives a 50 question multiple choice test with 4 choices for each question. What is the probability that Bart passes the test (60% correct)? P(x 30) = 50C30(.6 30 )(.4 20 ) + 50C31(.6 31 )(.4 19 ) + + 50C50(.6 50 )(.4 0 ) = 1 binomcdf(50,.6, 29) = 1.4390 =.5610 What is the probability that the first problem Bart gets correct is the 4th question? P(1st correct is 4th question) = (.4 3 ) (.6) = geompdf(.6, 4) =.0384 35/36
Example Let us stipulate that your good friend Bart Simpson is a 60% student. That is, he averages 60% on tests. His teacher gives a 50 question multiple choice test with 4 choices for each question. What is the probability that the first question Bart gets correct is in the first 4? P(1st correct 4) = (.4 0 )(.6)+ (.4 1 )(.6) + + (.4 3 )(.6) = geomcdf(.6, 4) =.9744 What is the probability that the first question Bart gets correct is after the 2nd? P(1st correct after 2nd) = 1 [(.4 0 )(.6)+ (.4 1 )(.6) = 1 geomcdf(.6, 2) =.16 36/36