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Solution here sample space is note that ns36 let aa

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Solution:Here sample space is:Note that n(S)=36Let A=a total of 5 occurs={(1,4), (2,3), (3,2), (4,1)}, n(A)=4, P(A)=4/36Let B= a total of 11 occurs={(5,6), (6,5)}, n(B)=2, P(B)=2/36Note that A & B are mutually exclusive events,So P(AUB)=P(A)+P(B)=4/36+2/36=6/36=1/6Example:Three horses A, B and C are in a race; A is twice as likely to win as B and B is aslikely to win as C what is the probability that A or B wins?Solution:Let P(C)=p then P(B)=2P(C)=2p and P(A)=2P(B)=2(2p)=4pSince A, B and C are mutually exclusive and collectively exhaustive events,So P(A)+P(B)+P(C)=1p+2p+4p=1,7p=1, orp=1/7So, P(C)=p=1/7,P(B)=2p=2/7, P(A)=4p=4/7P(A or B wins)= P(AUB)=P(A)+P(B)=4/7+2/7=6/7() () () () () ()() () () () () ()() () () () () ()() () () () () ()() () () () () ()() () () () () (){ 1,1 , 1,2 , 1,3 , 1,4 , 1,5 , 1,6 ,2,1 , 2,2 , 2,3 , 2,4 , 2,5 , 2,6 ,3,1 , 3,2 , 3,3 , 3,4 , 3,5 , 3,6 ,4,1 , 4,2 , 4,3 , 4,4 , 4,5 , 4,6 ,5,1 , 5,2 , 5,3 , 5,4 , 5,5 , 5,6 ,6,1 , 6,2 , 6,3 , 6,4 , 6,5 , 6,6 }S=
Lecture 19Lecture OutlineConditional probabilityIndependent and Dependent EventsRelated ExamplesConditional ProbabilityThe sample space for an experiment must often be changed when some additional informationrelated to the outcome of the experiment is received.The effect of such additional information is to reduce the sample space by excluding someoutcomes as being impossible which before receiving the information were believed possible.The probabilities associated with such a reduced sample space are called conditionalprobabilities.Example:Let us consider the die throwing experiment with sample space=S={1,2,3,4,5,6}Suppose we wish to know the probability of the outcome that the die shows 6, say event A. So,P(A)=1/6=0.166If before seeing the outcome, we are told that the die shows an even number of dots, say event B.Then this additional information that the die shows an even number excludes the outcomes 1,3and 5 and thereby reduces the original sample space to only three numbers {2,4,6}. SoP(6)=1/3=0.333We call 1/3 or 0.333 as the conditional probability of event A because it is computed under thecondition that the die has shown even number of dots.P(Die shows 6/die shows even numbers)=P(A/B)=1/3=0.333Example:Two coins are tossed. What the probability that two heads result, given that there is at-least one head?Solution:S={HH,HT,TH,TT},n(S)=4Let A=Two Heads appear={HH}Let B=at-least one head={HH,HT,TH}P(A/B)=?We have, P(A/B)=P(AB)/P(B)P(A)=1/4,P(B)=3/4(AB)={HH},P(AB)=1/4()()()()()()()()()()()/,0n ABn ABn SP ABP ABP Bn Bn BP Bn S===
P(A/B)=P(AB)/P(B)=(1/4)/(3/4)=1/3=0.33Example:Three coins are tossed. What the probability that two tails result, given that there is at-least one head?Solution:S={HHH,HHT,HTH, THH, HTT, THT, TTH, TTT}, n(S)=8Let A=Two tails appear={HTT, THT, TTH }Let B=at-least one head={HHH,HHT,HTH, THH, HTT, THT, TTH}P(A/B)=?We have, P(A/B)=P(AB)/P(B)P(A)=3/8,P(B)=7/8(AB)={HTT, THT, TTH },P(AB)=3/8P(A/B)=P(AB)/P(B)=(3/8)/(7/8)=21/64Example:A pair of dice is thrown, what is the probability that the sum of two dice will be 4,given that (i). The two dice has the same outcome.

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Term
Fall
Professor
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Tags
Mean, Frequency distribution, Ungrouped Data

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