Bernoulli Experiments, Binomial Distribution

1 Bernoulli Experiments, Binomial DistributionIf a person randomly guesses the answers to 10 multiplechoice questions, we can ask questions likeIwhat is the probability that they get none right?Iwhat is the probability that they get all ten right?Iwhat is the probability that they get at least threeright?Ihow many do they get right on average?These and similar scenarios lead toBernoulli Experimentsandthe Binomial Distribution . ABernoulli Experimentinvolves repeated (in this case 10) independent trials of anexperiment with 2 outcomes usually called success and failure (in this case getting a question right/wrong). Bernoulli experiment withnTrialsHere are the rules for a Bernoulli The experiment is repeated a fixed number of times (ntimes).2. Each trial has only two possible outcomes, success and failure . The possible outcomes are exactly thesame for each The probability of success remains the same for eachtrial. We usepfor the probability of success (on eachtrial) andq= 1 pfor the probability of The trials are independent (the outcome of previoustrials has no influence on the outcome of the nexttrial).

2 5. We are interested in the random variableXwhereX= the number of successes. Note the possible valuesofXare 0,1,2,3, .. , a coin 12 times, count the number of heads. Heren= 12. Each flip is a trial. It is reasonable to assume thetrials are independent. Each trial has two outcomes heads(success) and tails (failure). The probability of success oneach trial isp= 1/2 and the probability of failure isq= 1 1/2 = 1/2. We are interested in the variableXwhich counts the number of successes in 12 trials. This isan example of a Bernoulli experiment with 12 basketball player takes four independent free throws witha probability of of getting a basket on each shot. Thenumber of baskets made is recorded. Here each free throw isa trial and trials are assumed to be independent. Each trialhas two outcomes basket (success) or no basket (failure).The probability of success isp= and the probability offailure isq= 1 p= We are interested in the variableXwhich counts the number of successes in 4 trials.

3 This isan example of a Bernoulli experiment with 4 bag contains 6 red marbles and 4 blue marbles. Fivemarbles are drawn from the bagwithout replacementand the number of red marbles is observed. We might let atrial here consist of drawing a marble from the bag and letsuccess be getting a red. However, this isnota Bernoulliexperiment since the trials are not independent (the mix ofreds and blues changes on each trial since we do not replacethe marble) and the probability of success and failure varyfrom trial to bag contains 6 red marbles and 4 blue marbles. A marbleis drawn at random from the bag, its color is noted andthen it is replaced. Five marbles are drawn from the urn inthis way (withreplacement) and the number of redmarbles is observed. Thisisa Bernoulli experiment , whereeach time we draw a marble from the bag constitutes onetrial. Trials are independent since we draw randomly fromthe bag and the probability of success (getting a red) is thesame on each trial (p= 6/10) since we replace the marbleafter each draw.

4 We are interested in the number ofsuccesses in five trials of this people are chosen at random from among likelyDemocrat primary voters in Indiana, and asked do youplan to vote for Bernie Sanders in the upcoming primaryon May 3? . Their answers are recorded. Assuming peoplereally are randomly selected, this is an independentrepetition of the same trial 1000 times. If we declare votingfor Bernie success , thenp= the proportion of likelyDemocrat primary voters in Indiana who plan to vote forBernie. IfX= the number of successes, then a observationofXcan be used to estimate (unknown)p. For example, ifX= 442 in a particular running of this poll, we wouldconcluded thatpis likely around 44%.Probability Distribution ofXOur next goal is to calculate the probability Distribution forthe random variableX, whereXcounts the number ofsuccesses in a Bernoulli experiment withntrials. We willstart with a small example for which a tree diagram can bedrawn (we have already looked at a specific case of thisexample when we studied tree diagrams).

5 Example: A basketball player takes 4 independent freethrows with a probability of of getting a basket on eachshot. LetX= the number of baskets he that this is indeed a Bernoulli experiment withn= 4 andp= Distribution ofXUse the tree diagram below to find the probability that hegets exactly 2 baskets orP(X= 2).B = gets a basket, M = (X= 2) =C(4,2)( )2( )2= Distribution ofXIn general we have the following:IfXis the number of successes in a Bernoulliexperiment withnindependent trials, where theprobability of success ispin each trial (and the probabilityof failure is thenq= 1 p), thenP(X=k) =C(n, k)pkqn k=(nk)pkqn kfork= 0,1,2, , Distribution ofXWe can see why this is true if we visualize a tree diagramfor thenindependent trials. The number of paths withexactly k success (out ofntrials) isC(n, k) and theprobability of every such path equalspkqn k. The eventthatX=kcan result from any one of these outcomes(paths), hence theP(X=k) is the sum of the probabilitiesof all paths with exactlyksuccesses which isC(n, k)pkqn basketball player takes 4 independent free throws with aprobability of of getting a basket on each shot.

6 LetX=the number of baskets he gets. Write out the fullprobability Distribution (X)01234 ExampleXP(X)0C(4,0)( )0( ) (4,1)( )1( ) (4,2)( )2( ) (4,3)( )3( ) (4,4)( )4( ) + + + + = 1 Binomial Random VariablesFor a Bernoulli experiment withntrials, letXdenote thenumber of successes in thentrials, where the probability ofsuccess in each trial isp. This Distribution of random thevariableXis called abinomial Distribution with valueofXisE(X) =npand thestandard deviation ofXis (X) = npqwhereq= 1 a basketball player takes 8 independent free throws, witha probability of of getting a basket on each shot, whatis the probability that she gets exactly 6 baskets?C(8,6)( )6( )2 is the expected number of baskets that she gets?np= 8(.7) = (not6, or 5! expected value doesn thave to be a value that can actually occur)ExamplesA student is given a multiple choice exam with 10questions, each question with five possible answers.

7 Heguesses randomly for each question.(a) What sP(he will get exactly 6 questions correct)?n= 10,k= 6,p= , soC(10,6)( )6( )4 (b) What is the probability he will getat least6?C(10,6)( )6( )4+C(10,7)( )7( )3+C(10,8)( )8( )2+C(10,9)( )9( )1+C(10,10)( )10( )0 (c) What is the expected number of correct answers, andwhat s the standard deviation?E(X) =np= 10 = 2, (X) = npq= 10 = slightly off-topic exampleAssume that the Mets and the Royals are in the worldseries, that the Mets have a 3/5 chance of winning anygiven game, and that the games are independentexperiments. What is the probability of a 7 game series?Note 1: The world series isnota Bernoulli experiment !(number of games is not fixed in advance)Note 2A seven game series will occur only when eachteam wins 3 of the first 6 seven game series will occur whenever the Mets winexactly 3 of the first 6 games. The probability of this isC(6,3)( )3( )3 is also true that a seven game series will occur wheneverthe Royals win exactly 3 of the first 6 games.

8 Theprobability of this isC(6,3)( )3( )3 quality control exampleThe Everlasting Lightbulb company produces light bulbs,which are packaged in boxes of 20 for shipment. Tests haveshown that 4% of their light bulbs are defective.(a)What is the probability that a box, ready for shipment,contains exactly 3 defective light bulbs?C(20,3)( )3( )17 (b) What is the probability that the box contains 3 or moredefective light bulbs?1 (C(20,2)( )2( )18+C(20,1)( )1( )19+C(20,0)( )0( )20) 1 = can also compute the expected number of defectivebulbs,E(X) = 20 = , and the standard deviation, (X) = 20 exampleSuppose that the voting population in Utopia is 300 millionand 60% of the voting population intend to vote forMelinda McNulty in the next election. We take a randomsample of size 100 from the same voting population and askeach person chosen whether they will vote for MelindaMcNulty in the next election or not. LetXbe the numberof YESes in our sample.

9 The possible values ofX(thenumber of successes) are 0,1,2,3, , 100, p= , q= 1 p= Calculate thefollowing:(a) What isP(X= 60)? (b) What isP(X620)?(c) What isP(X >70)? (d) What isP(X <50)?(e) What isP(506X660)?Polling exampleP(X= 60) (X620) 10 16P(X >70) = 1 P(X670) (X <50) =P(X649) (506X660) =(P(X= 50) +P(X= 51) + +P(X= 60)) (506X660) =P(X660) P(X649) Zener TestZener CardsOne controversial test for ESP involves using a deck ofZener cards. This deck consists of 5 copies of the 5 cardsshown below. The tester (sender) shuffles the 25 cardsthoroughly, looks at the one on top of the deck and sends the information to the receiver (the person being testedfor ), without letting them see the card of process is repeated many times. Many of the earlyresults about the test were controversial because of flaws inhow the test was conducted and miscalculation ofprobabilities (seeZener Cards Skepticism).ESP Zener TestIn this online version:Zener Test, you have to guess 25cards, each of which is selected randomly by the computerprior to your guess.

10 The actual card will be shown afteryou click on your chosen symbol. To show evidence of ESP,you need to score at least 10 correct guesses (hits). If theselection of the card is random, this is a BernoulliExperiment with 25 trials and a probability ofp= ofsuccess (correct guess) in Zener Test(a) What is the expected number of correct answers in theabove test if the person taking the test is guessing?E(X) = 25 ( ) = 2.(b) What isP(X>10)? (c) Approximately how many students in a class of 100would you expect to get a score of 10 or greater on this testby randomly guessing?100 P(X>10) = 100 = , so close to Exam questionsRecall the notationC(n, k) =(nk)1An Olympic pistol shooter has a23chance of hitting thetarget at each shot. Find the probability that he will hitexactly 10 targets in a game of 15 shots.(a) 1 (1510)(23)10(13)5(b)(1510)(23)10(13)5(c) (23)10(d)(1510)(23)5(13)10(e) 1 -(13)5 Old Exam questions(a) 1 (1510)(23)10(13)5(b)(1510)(23)10(13)5(c) (23)10(d)(1510)(23)5(13)10(e) 1 -(13)5(b) is hits exactly 10 targets out of 15 shots (so is correctanswer)(d) is hits exactly 5 targets out of 15 shots (a) is does not hit exactly 10 targets out of 15 shots (c) is hits 10 targets in his first 10 shots (e) is does not miss all 5 of his first 5 shots Old Exam questionsA random variableXis the number of successes in aBernoulli experiment withntrials, each with a probabilityof successpand a probability of failureq.