Question

Exercise 1 While walking around Super Mario World, you encounter some enemies. Enemies pop up one-at-a-time. You may assume that these encounters are independent. There are 4 potential enemies you may run into. Enemy Probability 0.5 Koopa 0.3 Goombah 0.15 Chargin Chuck 0.05 Воо a) Suppose you encounter 10 enemies What is the probability that at least 1 is a Goombah? b) Suppose you encounter 10 enemies What is the probability that more than 8 are Koopas? c) What is the probability that you see your first Chargin Chuck after your tenth encounter? d) What is the probability that you see your first Boo on your twentieth encounter? e) What is the probability that you see your second Goombah on your tenth encounter? f) Suppose you encounter 8 enemies. What is the probability that it is 5 Koopas, 2 Goombas and 1 Boo?

Answer 1

Given, Enemy Koopa Goombah Chargin Chunk Probability (p) 0.50 0.30 0.15 0.05 a) Let X be the random variable denote number of encounters of Goombah, out of n=10 enemy encounters. Then X follows binomial distribution with parameters n,p Then probability mass function of X is PCX = x)=(*) p*(1-p)** ; X=0,1,2,....n we need to calculate P(X> 1) = 1- P(X<1)= 1-P(X=0) P(X =0) =( 10 )0.3°(1 – 0.3)" ° = 0,0282 So the required probability = 1-0.0282 = 0.9718 b) We need to calculate probability of more than 8 encounters of Koopas, out of 10 enemy encounters. By the similar arguments mentioned in part (a) we need to calculate P(X >8) PCX >8)= P(X =9) + P(X = 10) = 69 70.5*(1-0.5)*4 +G100.59 (1–0.5 )'-10 = 0.0098 +0.0010 -0.0108 c) let us define a SUCCESS as seeing a chargin Chunk and FAILURE as not seeing chargin chunk in particular encounter, let X denote the number of FAILURES before the First SUCCESS, then X follows geometric distribution with parameter p. Then probability mass function of X is P(X = x)=(1-p)*p; X = 0,1,2,... we need to calculate the probability of getting a success (seeing a charging chunk) after the 10th trial. ie Probability of exactly 10 failures before 1 success. ie P(x = 10) = (1 -0.15) 0.15 = 0.0295 d) By using the similar argument part (c) we need to calculate the probability of getting 19 failures before the first success. i.e P(X = 19)=(1-0.05) 0.05 = 0.0189

e) Again, let us define SUCCESS as the encounter is Goombah and Failure as the encounter is not Goombah. Let X denote the number of failures before the rth success. Then X follows Negative binomial ( a generalisation of geometric probability law) with parameters r, p Then probability mass function of X is P(X = x)=1***p" (1-p)*; X=0,1,2,3.... here we need to calculate probability of seeing second goombah at 10th encounter. ie. 10th encounter is goombah, and in the preceding 9 encounters, there are exactly 8 failures( not seeing goombah) and 1 success ( seeing goombah). So the required probability is (8+2–160.15) (1–0.15) P(X=8) = 8 /(0.15)(1-0 = 9 * 0.0225 * 0.2725 = 0.0552