Question

Audrey and Diana go fishing at the Lyndon Fishing Pond. Upon arrival the owner informs them that the pond is stocked with an infinite number of independent fish, and that a typical fisher catches fish at a Poisson rate of 2 fish per hour. There are 8 other people fishing there that day. Diana has the same skill level as a typical fisher but Audrey catches on average twice as many fish as a typical fisher.

For the rest of the question, assume that 100 fish were caught that day.

Use those rounded probabilities in parts b), c) and d):
i. The probability a fish was caught by Audrey is 0.182
ii. The probability a fish was caught by Diana is 0.091
iii. The probability a fish was caught by someone else is 0.727

(c) (2) Find the probability that Audrey catches 15 fish and Diana catches 15 fish

(d) (2) Find the probability that Audrey and Diana catch 30 fish together

(e) (2) Given that Audrey catches 15 fish, find the probability that Diana catches 15 fish

(f) (2) Explain logically the difference between the probabilities in (c), (d), and (e)

Answer 1

c.

probability that Audrey catches 15 fish and Diana catches 15 fish

= P(audrey catches 15 of 100 fishes)*P(diana catches 15 of 85 remaining fishes)

= [ 100C15*(P(audrey)^15)*(1-P(audrey))^(100-15) ] * [ 85C15*(P(diana)^15)*(1-P(diana))^(85-15) ]

= [ (2.5333847*10^17)*(P(audrey)^15)*(1-P(audrey))^(100-15) ] * [ ( 1.7984495*10^16 )*(P(diana)^15)*(1-P(diana))^(85-15) ]

= [ (2.5333847*10^17)*(0.182^15)*(1-0.182)^(100-15) ] * [ ( 1.7984495*10^16 )*(0.091^15)*(1-0.091)^(85-15) ]

= 0.000425

d.

P(audrey or diana) = P(audrey) + P(diana) = 0.273

P(audrey and diana catch 30 fish together) = 100C30*(P(audrey or diana)^30)*(1-P(audrey or diana))^(100-30)

= 100C30*( 0.273^30)*(1- 0.273)^(100-30)

= ( 2.937234*10^25 )*( 0.273^30)*(1- 0.273)^(100-30)

= 0.07248

e.

Given that Audrey catches 15 fish, find the probability that Diana catches 15 fish

= P(audrey catches 15 of 100 fishes)*P(diana catches 15 of 85 remaining fishes) / P(audrey catches 15 of 100 fishes)

{ P(audrey catches 15 of 100 fishes)*P(diana catches 15 of 85 remaining fishes) calculated in part c}

putting values form part c :

= 0.000425 / [ (2.5333847*10^17)*(0.182^15)*(1-0.182)^(100-15) ]

= 0.00549

f.

in (c) we calculate probability that each of audrey and diana catch 15 fishes each

in (d) we calculate probability that audrey and diana catch 30 fish together which doesn't imply any fixed value on each

in (e) we are given audrey catched 15 fishes so we are calculating probability that of the remaining 85 fishes audrey caught 15 fishes

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