Download (1) There are two coins. The first one is fair while the second lands

(1) There are two coins. The first one is fair while the second lands
on head 60 percent of the time. One coin is selected at random
and tossed three times.
(a) Find the probability that 2 heads and one tail appear;
(b) If two heads and one tail had appeared, what is the probability that the next toss will be head?
Solution. Let C1 be the event that the first coin is chosen
and C1 be the event that the second coin is chosen.
(a) We have P (2H, 1T |C1) = 3(0.5)3 P (2H, 1T |C2) = 3(0.6)2 (0.4),
so by the Law of Total Probability the answer is
1
1
3(0.5)3 + 3(0.6)2 (0.4).
2
2
(b) By Bayes Formula
P (C1|2H, 1T ) =
1
3(0.5)3
2
1
3(0.5)3 + 12 3(0.6)2 (0.4)
2
=
125
,
269
P (C2|2H, 1T ) =
1
3(0.6)2 (0.4)
2
1
3(0.5)3 + 12 3(0.6)2 (0.4)
2
=
144
.
269
By the Law of Total Probability
P (H|2H1T ) =
125 1 144 6
1489
+
=
.
269 2 269 10
2690
(2) To check for a certain illness two tests are used. If the person
is ill the first test gives positive result 99% of times but it gives
false positive 10% of times. The second test gives positive result
90 % of times when the person is ill and it gives a false positive
3% of times. It is known that 10% of patients have the illness.
(a) Suppose only one test is performed and comes positive.
In which case the patient is more likely to be ill: if they took
test 1 or test 2?
(b) Suppose that the outcomes of two tests are independent.
If both tests come positive how like the patient to be ill?
Solution. Let I be the event that the patient is ill and P1 , P2
be the event that test 1, respectively test 2, is positive.
(a) By Bayes Fomula
P (I|P1 ) =
(0.1)(0.99)
≈ 0.524,
(0.1)(0.99) + (0.9)(0.1)
P (I|P2 ) =
(0.1)(0.9)
≈ 0.769.
(0.1)(0.9) + (0.9)(0.03)
1
2
(b) We have
P (P1 P2 |I) = (0.99)(0.9),
P (P1 P2 |I c ) = (0.1)(0.03)
so by Bayes Formula
P (I|P1 P2 ) =
(0.1)(0.99)(0.9)
≈ 0.971.
(0.1)(0.99)(0.9) + (0.9)(0.1)(0.03)
(3) Three balls are drawn from three urns. The first urn contains 1
blue and 5 red balls, the second urn contains 2 blue and 4 red
balls, and the third urn contains 3 red and 3 green balls.
(a) Find the probability that 2 red balls are chosen;
(b) Let X be the number of different colors chosen. Find the
distribution of X.
Solution. (a) Considering three cases depending on which
ball is not read we get
143 523 543
102
P (2 reds) =
+
+
=
.
666 666 666
216
(b) If there is only one color then all balls should be red.
Thus
543
60
P (X = 1) =
=
.
666
216
If there are three color then the last ball must be green. Considering whatever first or second ball are red we get
42
523 143
+
=
.
P (X = 3) =
666 666
216
Hence
114
P (X = 2) = 1 − P (X = 1) − P (X = 3) =
.
216
(4) 10000 bacteria are analyzed in the lab. It is known that the
probability that a bacteria has gene A is 21 and the probability
1
that it has gene B is 5000
. Compute approximately the probability that
(a) 5010 or more bacteria will carry gene A;
(b) Exactly 3 bacteria care gene B.
Solution. Let X be the number of bacteria carrying gene A
and Y be the number of bacteria carrying gene B.
(a) X ≈ N (5000, 2500), that is X ≈ 5000 + 50Z where Z is
the standard normal. Accordingly
P (X ≥ 5010) ≈ P (Z ≥ 0.2) = 1 − P (Z ≤ 0.2) ≈ 0.42.
(b) Y ≈ Pois(2). Accordingly P (Y = 3) ≈
23 −2
e
3!
≈ 0.18.
3
(5) Jane finds a job which requires her to commute 5 days a week.
1
On her way home Jane is in a harry so there is 20
probability
that she gets a speeding ticket
(a) Let X be the number of tickets Jane gets during first 6
weeks of work. Compute EX and V X.
(b) When Jane gets three tickets she needs to attend a driving
school. Find the probability that Jane gets her third ticket on
her 50th commute.
1
). Therefore
Solution. (a) X ∼Bin(30, 20
EX = 30 ×
1
3
= ,
20
2
VX =
30 1
19
57
×
= .
× 20 20
40
(b) Using the formula for negative binomial distribution with
1
parameters 3 and 20
we see that the answer is
47 3
19
1
49
.
2
20
20
(6) A class has 15 boys and 20 girls. 10 theater tickets are distributed at random.
(a) Find the probability that girls have exactly 6 tickets;
(b) Amanda’s lunch mates are Barbara, Cindy, Dalia and
Elena. Find the conditional probability that Amanda’s table
gets exactly 2 tickets given that girls got exactly 6 tickets.
Solution. (a) From the formula for hypergeometric distribution we see that the answer is
15
20
4
6
.
35
10
(b) Given that the girls get 6 tickets the number of tickets
obtained by Amanda’s table is hypergeometric with parameters
(20, 5, 6). So the answer is
5
15
2
4
.
20
6
(7) A number of misprints on a page has Poisson distribution with
parameter 12 .
4
(a) Find the probability that exactly three of the next 10
pages will have at least two misprints.
(b) Let X be the first page which has a misprint. Find EX
and V X.
Solution. (a) The probability that a page has no misprints is
1
√1 , the probability that a page has one misprint is √
, therefore
e
2 e
the probability that a page has two or more misprints is 1− 2√3 e .
Using the formula for binomial distribution we see that the
answer is
7 3
3
3
10
√
1− √
.
3
2 e
2 e
(b) X has geometric distribution with parameter 1 − √1e . Accordingly
√
√
1 √
EX = e, V X = (1 − √ )( e)2 = e − e.
e
(8) Let X1 have density equal to c1 x3 on [0, 1] and zero elsewhere
and X2 have density equal to c2 x10 on [0, 2] and zero elsewhere.
(a) Compute c1 and c2 ;
(b) Which of the two random variables above has a smaller
variance?
Solution. (a) We have
−1
Z 2
−1
Z 1
11
10
3
= 11 .
x dx
= 4, c2 =
x dx
c1 =
2
0
0
(b) We have
Z 1
Z 1
4
2
3
2
EX1 =
4x dx = , E(X1 ) =
4x5 dx = .
5
3
0
0
Hence
2
2
4
V X1 = −
≈ 0.0267.
3
5
On the other hand
Z
Z
11 2 11
212 11
11
11 2 12
213 11
44
2
x dx =
=
,
E(X
)
=
x
dx
=
=
.
EX2 = 11
2
2
12 211
6
211 0
13 211
13
0
Hence
2
44
11
11
V X2 =
−
=
≈ 0.0235.
13
6
468
Thus V X2 < V X1 .
5
(9) The lifetime of a light bulb (measured in days) has exponential
distribution with parameter 1/100.
(a) Find the distribution of the lifetime measured in hours;
(b) If the bulb is installed on a Wednesday at noon, find the
probability that it will burn out on a Monday.
Solution. (a) Let X be the lifetime in days and Y be the life1 1
1
time in hours. Then Y = 24X hence Y ∼Exp( 100
) =Exp( 2400
).
24
(b) Considering each Monday separately we get
∞
X
P (Burn out on Monday) =
P (4.5 + 7k ≤ X ≤ 5.5 + 7k) =
k=0
∞
X
k=0
[e−0.045−0.07k − e−0.045−0.07k ] =
e−0.045 − e−0.055
.
1 − e−0.07
(10) The amount of sales at a department store on a given day has
normal distribution with mean 30000 and standard deviation
3000. Find the probability that the store sold
(a) more than 31000 worth of goods;
(b) between 28000 and 32000 worth of goods.
Solution. Let X be the amount of sales. Then X = 30000 +
3000Z where Z is the standard normal random variable. Thus
1
1
(a) P (X > 31000) = P (Z > ) = 1 − P (Z < ) ≈ 0.37,
3
3
2
2
(b) P (2800 < X < 3200) = P (− < Z < ) =
3
3
2
2
2
P (Z < ) − P (Z < − ) = 2P (Z < ) − 1 ≈ 0.50.
3
3
3