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Question 3: My dog Peter lies to lick Jack, his cat companion. Because Jack prefers not to be Ecked by a slobbery dog, be tries to dodge Peters tongue, so that the licks only hit their intended arget with probability 0.7, independently. One afternoon, Peter aims 8 licks at the cat. (a) (5 pts) What is the probability that Peter lands exactly 4 licks on Jack? (b) (10 pts) Given that he lands at least two licks, what is the probability that Peter lands exactly 4 licks?
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Answer #1

a)

P(Peter lands exactly 4 licks)=P(X=4)=8C4(0.7)4*(0.3)4 =0.1361

b)

P(at least 2 licks)=P(X>=2) =1-P(X<=1)=1-(P(X=0)+P(X=1))=1-( 8C0(0.7)0*(0.3)8 +8C1(0.7)1*(0.3)7 )

=0.9987

hence P(Peter lands exactly 4 licks given at least 2 licks)=P(X=4|X>=2)=0.1361/0.9987=0.1363

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