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3.1 Problems on conditional probability

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Given the following data:

P ( A ) = 0 . 55 , P ( A B ) = 0 . 30 , P ( B C ) = 0 . 20 , P ( A c ∪ B C ) = 0 . 55 , P ( A c B C c ) = 0 . 15

Determine, if possible, the conditional probability P ( A c | B ) = P ( A c B ) / P ( B ) .

% file npr03_01.m % Data for [link] minvec3 DV = [A|Ac; A; A&B; B&C; Ac|(B&C); Ac&B&Cc];DP = [ 1 0.55 0.30 0.20 0.55 0.15 ];TV = [Ac&B; B];disp('Call for mincalc') npr03_01Variables are A, B, C, Ac, Bc, Cc They may be renamed, if desired.Call for mincalc mincalcData vectors are linearly independent Computable target probabilities1.0000 0.2500 2.0000 0.5500The number of minterms is 8 The number of available minterms is 4- - - - - - - - - - - - P = 0.25/0.55P = 0.4545
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In Exercise 11 from "Problems on Minterm Analysis," we have the following data: A survey of a represenative group of students yields the following information:

  • 52 percent are male
  • 85 percent live on campus
  • 78 percent are male or are active in intramural sports (or both)
  • 30 percent live on campus but are not active in sports
  • 32 percent are male, live on campus, and are active in sports
  • 8 percent are male and live off campus
  • 17 percent are male students inactive in sports

Let A = male, B = on campus, C = active in sports.

  • (a) A student is selected at random. He is male and lives on campus. What is the (conditional) probability that he is active in sports?
  • (b) A student selected is active in sports. What is the(conditional) probability that she is a female who lives on campus?
npr02_11 - - - - - - - - - - - - mincalc- - - - - - - - - - - - mincalctEnter matrix of target Boolean combinations [A&B&C; A&B; Ac&B&C; C] Computable target probabilities1.0000 0.3200 2.0000 0.44003.0000 0.2300 4.0000 0.6100PC_AB = 0.32/0.44 PC_AB = 0.7273PAcB_C = 0.23/0.61 PAcB_C = 0.3770
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In a certain population, the probability a woman lives to at least seventy years is 0.70 and is 0.55 that she will live to at least eighty years.If a woman is seventy years old, what is the conditional probability she will survive to eighty years? Note that if A ⊂ B then P ( A B ) = P ( A ) .

Let A = event she lives to seventy and B = event she lives to eighty. Since B ⊂ A , P ( B | A ) = P ( A B ) / P ( A ) = P ( B ) / P ( A ) = 55 / 70 .

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From 100 cards numbered 00, 01, 02, ⋯ , 99, one card is drawn. Suppose A i is the event the sum of the two digits on a card is i , 0 ≤ i ≤ 18, and B j is the event the product of the two digits is j . Determine P ( A i | B 0 ) for each possible i .

B 0 is the event one of the first ten is drawn. A i B 0 is the event that the card with numbers 0 i is drawn. P ( A i | B 0 ) = ( 1 / 100 ) / ( 1 / 10 ) = 1 / 10 for each i , 0 through 9.

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Two fair dice are rolled.

  1. What is the (conditional) probability that one turns up two spots, given they show different numbers?
  2. What is the (conditional) probability that the first turns up six, given that the sum is k , for each k from two through 12?
  3. What is the (conditional) probability that at least one turns up six, given that the sum is k , for each k from two through 12?
  1. There are 6 × 5 ways to choose all different. There are 2 × 5 ways that they are different and one turns up two spots. The conditional probability is 2/6.
  2. Let A 6 = event first is a six and S k = event the sum is k . Now A 6 S k = ∅ for k ≤ 6 . A table of sums shows P ( A 6 S k ) = 1 / 36 and P ( S k ) = 6 / 36 , 5 / 36 , 4 / 36 , 3 / 36 , 2 / 36 , 1 / 36 for k = 7 through 12, respectively. Hence P ( A 6 | S k ) = 1 / 6 , 1 / 5 , 1 / 4 , 1 / 3 , 1 / 2 , 1 , respectively.
  3. If A B 6 is the event at least one is a six, then P ( A B 6 S k ) = 2 / 36 for k = 7 through 11 and P ( A B 6 S 12 ) = 1 / 36 . Thus, the conditional probabilities are 2/6, 2/5, 2/4, 2/3, 1, 1, respectively.
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Source:  OpenStax, Applied probability. OpenStax CNX. Aug 31, 2009 Download for free at http://cnx.org/content/col10708/1.6
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