302.102 Math Fall 2010 Solutions to Assignment #3 Let A be the event that a randomly selected person is a man so that Ac is the event that a randomly selected person is a woman. Let B be the event that a person is colour blind. We are told that P {B | A} = 0.05 and P {B | Ac } = 0.0025. (a) If men and women each make up the same proportion of the population, then P {A} = 0.5 and P {Ac } = 0.5, so that Bayes Rule tells us that P {A | B } = P {B | A} P {A} P {B | A} P {A} (0.05)(0.5) = = c } P {Ac } P {B } P {B | A} P {A} + P {B | A (0.05)(0.5) + (0.0025)(0.5) 20 . = = 0.952381. 21
1.
(b) If there are twice as many women as men in this population, then P {A} = 1/3, P {Ac } = 2/3, and Bayes Rule tells us that P {A | B } = P {B | A} P {A} P {B | A} P {A} (0.05)(1/3) = = c } P {Ac } P {B } P {B | A} P {A} + P {B | A (0.05)(1/3) + (0.0025)(2/3) 10 . = = 0.909091. 11
Let A be the event that a randomly selected chip is good so that Ac is the event that a randomly selected chip is bad. Let B be the event that a chip passes the cheap chip test. We are told that P {A} = 0.8 and P {Ac } = 0.2. Since all good chips pass the cheap chip test, P {B | A} = 1, but since 10% of bad chips also pass the cheap chip test, we have P {B | Ac } = 0.1. (a) Using Bayes Rule we nd P {A | B } = P {B | A} P {A} (1)(0.8) P {B | A} P {A} = = c } P {Ac } P {B } P {B | A} P {A} + P {B | A (1)(0.8) + (0.1)(0.2) 40 . = = 0.975609. 41
2.
(b) If the company sells all chips which pass the cheap chip test, then the percentage of chips sold that are bad is simply 1 40 P {Ac | B } = 1 P {A | B } = 1 =. 41 41
Let A be the event that a randomly selected person has the disease so that Ac is the event that a randomly selected person does not have the disease. Let B be the event that that the laboratory test on the blood sample returns a positive result. We are told that P {A} = 0.01 so that P {Ac } = 0.99. We are also told that P {B | A} = 0.95 and P {B | Ac } = 0.02. Using Bayes Rule we nd P {A | B } = P {B | A} P {A} (0.95)(0.01) P {B | A} P {A} = = c } P {Ac } P {B } P {B | A} P {A} + P {B | A (0.95)(0.01) + (0.02)(0.99) 95 . = = 0.324232. 293
3.
4.
The key to solving this problem is to realize that the event A from the previous problem needs to change. No longer is the patient who walks into the doctors oce randomly selected from the population. Hence, we take A to be the event that the patient has the disease. The doctors opinion is that P {A} = 0.3. If the blood test result is positive, then Bayes Rule implies P {A | B } = P {B | A} P {A} P {B | A} P {A} (0.95)(0.30) = = c } P {Ac } P {B } P {B | A} P {A} + P {B | A (0.95)(0.30) + (0.02)(0.70) 285 . = 0.953177. = 299
5.
For j = 1, 2, . . ., let Aj be the event that a 3 appears on roll j , let Bj be the event that a 1 or a 6 appears on roll j , let Cj be the event that a 2 appears on roll j , let Dj be the event that a 4 appears on roll j , and let Ej be the event that a 5 appears on roll j . If W denotes the event that Player A wins, then the Law of Total Probability implies P {W } = P {W | A1 } P {A1 } + P {W | B1 }P {B1 } + P {W | C1 } P {C1 } + {W P | D1 } P {D1 } + P {W | E1 } P {E1 } Since Player A wins immediately if a 3 appears on the rst roll, P {W | A1 } = 1, and since Player A loses immediately if a 1 or a 6 appears on the rst roll, P {W | B1 } = 0. Furthermore, since the die is fair, P {A1 } = P {C1 } = P {D1 } = P {E1 } = 1/6. Therefore, P {W } = 11 1 1 + P {W | C1 } + P {W | D1 } + P {W | E1 } . 66 6 6
The next observation is that P {W | C1 } = P {W | D1 } = P {W | E1 }. If Player A rolls a 2 initially, then the only way for Player A to win is if a 2 is rolled before a 3 on subsequent rolls. Similarly, if Player A rolls a 4 initially, then the only way for Player A to win is if a 4 is rolled before a 3 on subsequent rolls, and if Player A rolls a 5 initially, then the only way for Player A to win is if a 5 is rolled before a 3 on subsequent rolls. Since the die is fair, all of these events have the same probability. Thus, 11 P {W } = + P {W | C1 } . 62 The remaining step is to compute P {W | C1 }. It is reasonable to guess that P {W | C1 } = 1/2. This is because if a 2 is rolled initially, then the only way for Player A to win is if a 2 appears before a 3 on subsequent rolls. Since only a 2 or 3 matter (all other numbers cause a re-roll), and since 2 and 3 are both equally likely, we must have P {W | C1 } = 1/2. Alternatively, we can sum up an innite series as follows. (continued)
Let Fj = Bj Dj Ej be the event that 1, 4, 5, or 6 appears on roll j . Then, using the fact that the results of subsequent rolls are independent, we nd P {W | C1 } = P {C2 } + P {F2 } P {C3 } + P {F2 } P {F3 } P {C4 } + = = = 1 41 ++ 6 66 2 1 1+ + 6 3 4 6 2 3
2
2
1 + 6
+
1 1 6 1 2/3 1 =. 2 P {W } = 5 1 11 +=, 6 22 12 5 7 =. 12 12
This implies that
and so the probability that Player A loses is P {W c } = 1
6.
(a) Since
1
x2 dx = x1
1
= 0 (1) = 1
we see that taking c = 1 makes f a legitimate probability density. (b) Since
1
x1 dx = ln(x)
1
=0=
we see that there is no such c that makes f a legitimate probability density. (c) Using integration-by-parts twice (see Prerequisite Review Handout) gives x2 ex dx = x2 ex 2xex 2ex . Therefore, since
1 1 1
xe
2 x
dx = x e
2 x
2xe
x
2e
x 1
= [e1 2e1 2e1 ] [e1 +2e1 2e1 ] = e 5e1 ,
we see that taking c = (e 5e1 )1 = e/(e2 5) makes f a legitimate probability density. (d) As in (c), using integration-by-parts twice gives
0
x2 ex dx = x2 ex 2xex 2ex
0
= 0 (2) = 2
and so we see that taking c = 1/2 makes f a legitimate probability density.
(e) Since
0 0
xex dx = [xex ex ]
= (0 1) (0 0) = 1
by see that taking c = 1 makes f a legitimate probability density. Note that xex 0 for x 0. This means that if we multiply xex by a negative number it will always be non-negative. Hence, f (x) = xex for x 0 is a non-negative function that integrates to 1. (f ) Note that the function xex assumes both positive and negative values when 1 x 1. This means that there is no single value of c that could be multiplied by xex to make it strictly non-negative. Hence, there is no possible value of c that makes f a legitimate probability density.