How can i calculate the probability of obtaining a building permit?(really urgent for an upcoming test)

hello,

i was told that a variation of this question will be on an upcoming test, and i don't know how to solve it. can anyone please solve it and explain how he did it?

question:

in the first discussion of a municipal committee regarding illegal buildings:

  • the probability that the committee will order that building's destruction is 0.5
  • the probability that they will choose to do a second discussion on the building is 0.4.
  • the probability they will give it a permit making it legal is 0.1.

if in the first discussion the committee orders the destruction of a building, the owner submits an appeal(objection) in a probability of 0.7.

  • the probability the appeal (objection will be accepted) and the building will get a permit is 0.4
  • the probability the appeal will be denied and the building will be destroyed is 0.6

if on the first discussion the committee assigns a date for the second discussion regarding a building, the probability it will get a permit is 0.8. otherwise it is destroyed

the committee discusses the future of 20 illegal buildings, and there's no statistical relation between one decision to another.

eventually, each building either gets a permit or is destroyed.

1)what is the probability that the 15th building that the committee will discuss, will be the 2nd (among 15th) that will get a permit on the first discussion?

2)if eventually, 14 out of 20 buildings got a permit, what is the probability that 3 of them will get the permit on the first discussion?

3)if known that only 3 out of 20 buildings got a permit on the first discussion, what is the probability function of the additional building(out of 20) that got a permit eventually(meaning, after the first discussion)?

please help me with that. knowing how to solve it correctly will help me so much on my upcoming test

1 Answer
MattyMatty
Jun 2, 2018

#"1) 0.0356"#
#"2) 0.2381"#
#"3) (C(17,k) * 0.46^k * 0.44^(17-k)) / 0.9^17#

Explanation:

#P["permit"] = 0.1 + 0.4*0.8 + 0.5*0.7*0.4 = 0.56#
#P["permit on 1st disc."] = 0.1#

#"1) "= P["1 permit on first disc. in the first 14"]*P["15th permit on 1st disc."]#

#= C(14,1) * 0.1 * 0.9^13 * 0.1#
#= 0.0356#

#"2) A = 14 out of 20 buildings got a permit"#
#" B = 3 got a permit on first disc."#
#P[B|A] = (P[A" and "B]) / (P[A])#
#P[A] = C(20,14) 0.56^14 0.44^6 = 0.083894#
#P[A" and "B] = C(20,11,3) 0.46^11 0.1^3 0.44^6#
#= ((20!)/((11!)(3!)(6!))) 0.46^11 0.1^3 0.44^6#
#= 0.019977#
#=> P[B|A] = 0.019977/0.083894 = 0.2381#

#"3) For the other 17 buildings, all we know is that they did"#
#"not have a permit on first discussion."#
#"So they had a permit with probability (0.56 - 0.1)/0.9 or no"#
#"permit with probability 0.44/0.9."#
#=> P["k extra buildings got a permit"] #
#= C(17,k) (0.46/0.9)^k (0.44/0.9)^(17-k)#
#= (C(17,k) * 0.46^k * 0.44^(17-k)) / 0.9^17#

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