PROBABILITY PART B.pptx

# We will calculate p a 3 using the multiplication rule

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We will calculate P ( A 3 ) using the multiplication rule: P ( A 3 ) = P ( A 1 ∩ A 2 ∩ A 3 ) = P ( A 1 ) P ( A 2 | A 1 ) P ( A 3 | A 1 ∩ A 2 ) = We have P ( A 1) = 12/15

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since there are 12 student slots in groups other than the one of student 1, and there i.e. 3grad. & 12 undergrad are 15 student slots overall, excluding student 1. Similarly, P ( A 2 | A 1) = 8/14 , (2 grad.& 12 undergrad.) since there are 8 student slots in groups other than the one of students 1 and 2, and there are 14 student slots, excluding students 1 and 2. Also, P ( A 3 | A 1 ∩ A 2) = 4/13 (1 grad & 12 undergrad) , since there are 4 student slots in groups other than the one of students 1, 2, and 3, and there are 13 student slots, excluding students 1, 2, and 3.
Thus, the desired probability is P ( A 3 ) = P ( A 1 ∩ A 2 ∩ A 3 ) = P ( A 1 ) P ( A 2 | A 1 ) P ( A 3 | A 1 ∩ A 2 ) = 12/15 ·X 8/14 X · 4/13 = 0.141 and is obtained by multiplying the conditional probabilities

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Example. Radar detection. If an aircraft is present in a certain area, a radar correctly registers its presence with probability 0.99. If it is not present, the radar falsely registers an aircraft presence with probability 0.10. We assume that an aircraft is present with probability 0.05. What is the probability of false alarm (a false indication of aircraft presence), and the probability of missed detection (nothing registers, even though an aircraft is present)?
Let A and B be the events A = { an aircraft is present }, B = { the radar registers an aircraft presence }, and consider also their complements Ac = { an aircraft is not present }, Bc = { the radar does not register an aircraft presence }. The given probabilities are recorded along the corresponding branches of the tree describing the sample space, as shown in the tree diagram shown.

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Each event of interest corresponds to a leaf of the tree and its probability is equal to the product of the probabilities associated with the branches in a path from the root to the corresponding leaf. The desired probabilities of false alarm and missed detection are P (false alarm) = P ( Ac ∩ B ) = P ( Ac ) P ( B | Ac ) = 0 . 95 X · 0 . 10 = 0 . 095 , P (missed detection) = P ( A ∩ Bc ) = P ( A ) P ( Bc | A ) = 0 . 05 X 0 . 01 = 0 . 0005 .

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Example. We toss a fair coin three successive times. We wish to find the conditional probability P ( A | B ) when A and B are the events A = { more heads than tails come up }, B = { 1st toss is a head }. The sample space consists of eight sequences, Ω = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}, which we assume to be equally likely. The event B consists of the four elements HHH, HHT, HTH, HTT , so its probability is P ( B ) = 4/8 The event A ∩ B consists of the three elements outcomes HHH, HHT, HTH , so its probability is P ( A ∩ B ) = 3 / 8 . Thus, the conditional probability P ( A | B ) is P ( A | B ) = P ( A ∩ B ) / P ( B ) = (3 / 8 )/( 4 / 8) = 3 / 4
OR Because all possible outcomes are equally likely here, we can also compute P ( A | B ) using a shortcut.

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