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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!(103)! (.21)^3 (1.21)^103
n=10
x=3
p=chance (.21)
10! = 1*2*3*4… 10
3!=1*2*3
0!=1
p(e1)=1p(e)
P(E)=1P(E1)
P(X<=3)=P(0)+P(1)+P(2)+P(3)
P(X) = N!/X!(NX)! * pX(1P) ^ NX
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(1p)^x1
P(5)= (0.19)(10.19)^51
= (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)=1P(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!(NX)!
*
p^X(1p)^NX
P(X=4)=P(4)=12!/4!(124)!(.10)^4(1.1)^124
=1*2*3…*11*12/(1*2*3*4)(1*2*3…*8) *
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This note was uploaded on 12/09/2010 for the course BUS 2313 taught by Professor Zhang during the Fall '10 term at East Central.
 Fall '10
 Zhang
 Business

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