Question

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 Peter's 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?

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