BSEC Pg 1 - 1 total tirals is 10 2 probability of csuccess...

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1. total tirals is 10 2. probability of csuccess is 21% 3. Let x = #of successes Part A. P(X=3)=? P(X=3)=P(3) 10! / 3!(10-3)! (.21)^3 (1-.21)^10-3 n=10 x=3 p=chance (.21) 10! = 1*2*3*4… 10 3!=1*2*3 0!=1 p(e1)=1-p(e) P(E)=1-P(E1) P(X<=3)=P(0)+P(1)+P(2)+P(3) P(X) = N!/X!(N-X)! * pX(1-P) ^ N-X X= the trial number on which the first defective PC is found 1. the possibility of asuccess of a single call is p=0.19 2. Let X=the trial number on which the first sale occurs 3. the probability that the “first sale” occurs on trial X is given by P(X)=p(1-p)^x-1 P(5)= (0.19)(1-0.19)^5-1 = (0.19)(0.81)^4 = (0.19)(0.4305) = 0.082 (8.2%) P(X<=3) =P(1)+P(2)+P(3) P(x>3)=1-P(X<=3) 1-.4686=.5314 (53.14%) 1. n=12 2. p=.10 3. let x= # of people in the survey who prefer the cookie x is a RV with possible values the probability that there are exactly x “successes” among the n=12 trials is P(X)=N!/X!(N-X)! * p^X(1-p)^N-X P(X=4)=P(4)=12!/4!(12-4)!(.10)^4(1-.1)^12-4 =1*2*3…*11*12/(1*2*3*4)(1*2*3…*8) * =495(.0001)(.4305) =.0213 B. P(X>=4)=P(4)+P(5)…P(12) P(X>=4)=1-P(<4)x P(X>=4)=1-P(X<4) =1-.9742=0.0258 1. n=10 2. p=.23 3. Let x=# of students in the survey who use the card due to the rewards.
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