b)
(0.85)
9
= 0.232
The probability that the 11 a.m. flight takes off on time is 0.2316 (23.16%).
c)
1 – (0.85)
14
= 0.8972
The probability that the 1:30 p.m. flight is delayed is 0.8972 (89.72%).
7. a)
30
C
2
x
(0.05)
2
x
(0.95)
28
=
0.258636738
b) 1 P(2) – P1(1) – P(0)
1 –
30
C
2
x
(0.05)
2
x
(0.95)
28

30
C
1
x
(0.05)
1
x
(0.95)
29

(0.95)
30
= 0.187821187
8. a) I have created this table to help me solve this question.
Shot Number
Attempt #
Probability
1
Person A, 1
st
Attempt
0.166666667
2
Person B, 1
st
Attempt
0.138888889
3
Person A, 2
nd
Attempt
0.115740741
4
Person B, 2
nd
Attempt
0.096450617
5
Person A, 3
rd
Attempt
0.080375514
6
Person B, 3
rd
Attempt
0.066979595
7
Person A, 4
th
Attempt
0.055816329
8
Person B, 4
th
Attempt
0.046513608
9
Person A, 5
th
Attempt
0.03876134
10
Person B, 5
th
Attempt
0.032301117
11
Person A, 6
th
Attempt
0.026917597
12
Person B, 6
th
Attempt
0.022431331
13
Person A, 7
th
Attempt
0.018692776
14
Person B, 7
th
Attempt
0.015577313
15
Person A, 8
th
Attempt
0.012981094
16
Person B, 8
th
Attempt
0.010817579
17
Person A, 9
th
Attempt
0.009014649
18
Person B, 9
th
Attempt
0.007512207
19
Person A, 10
th
Attempt
0.006260173
20
Person B, 10
th
Attempt
0.005216811
In the table above, if Person A goes first, he would take shots 1,3,5,7,9,11,13, 15,17, and
19. If Person B goes second, he would take shots 2,4,6,8,10,12,14, 16, 18, and 20.
Since the cartridge is spun before every shot attempt,
the probability of "success" p is
the same for each outcome.
Since we are looking at
the number of trials required to
obtain the first hit,
we can use the geometric distribution formula to help us determine
whether or not it would be advantage to go second. By looking at the graph below, we
can deduce a general trend, which indicates that the probability of Person A getting hit
on their collective “attempts” is slightly higher than Person B. This is because the
probability of getting hit by the first shot on the first attempt is the greatest, and each
succeeding attempt has a slightly lower probability than the previous. For example, the
probability of the first hit being on the third shot (Person A) would be much higher than
the probability of the first hit being on the fourth shot (Person B). Therefore, the
probability of Person A getting hit first is higher than the probability that Person B gets hit
first. This pattern is characteristic of any geometric distribution. To prove this, we can add
up the probabilities of the attempts and determine whether Person A or Person B has a
higher chance of being shot before their tenth attempt. If we add up the probabilities of
Person A’s ten attempts (1,3,5,7,9,11,13, 15,17, and 19), we get a total probability value
of 0.53122688. Similarly, if we add up the probabilities of Person B’s ten attempts
(2,4,6,8,10,12,14, 16, 18, and 20), we get a total probability value of 0.442689067.
Therefore, by comparing the two values, we can see that it would clearly be an
advantage to take the second turn in ThumbTack Russian Roulette. I created the graph
below to visual represent this.
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 Fall '12
 jp
 Probability, Probability theory, Binomial distribution, Geometric distribution, Hypergeometric Distribution