1 Calculate the probability of winning the Irish lotto. Sol

```1
Calculate the probability of winning the Irish lotto.
Sol
Step one, calculate the number of combinations of choosing 6 numbers from 45.
The odds of winning are 1/8145060.
2
Consider the domain of dealing 5-card poker hands from a standard deck of 52
cards, under the assumption that the dealer is fair.
(a)
How many 5-card hands are there?
Sol
There are
(b)
five-card hands.
What is the probability of getting one of the 5-card hands?
Sol
1/
(c)
What is the probability of being dealt a royal straight flush (i.e. being dealt a hand
containing Ace, King, Queen, Jack and 10 all from the one suit)?
Sol
There are 4 possible hands that make up a royal straight flush. Therefore the probability of
getting a royal straight flush 4/2,598,960
What is the probability of being dealt a four of a kind (i.e. four kings, or four nines
etc.?
(d)
possible
Sol
There are 13 possible kinds and for each, the fifth card can be one of 48 possible other
cards. The total probability is therefore
3
A particular long traffic light in your morning commute is green 20% of the time that
you approach it. Assume that each morning represents an independent trail.
(a)
Over five mornings, what is the probability that the light is green on exactly one days?
Sol
P( X  1) 
(b)
Over 20 mornings, what is the probability that the light is green on exactly four days?
Sol
P( X  4) 
(c)
Over 20 mornings, what is the probability that the light is green on exactly more than
four days?
Sol
P( X  4)  1  P( X  4)  1  0.630  0.370
4
A multiple choice test contains 20 questions, each with four answers. Assume a student
just guesses on each question.
 0.2 0.8
5
1
1
 0.410
4
 0.2 0.8
20
4
4
16
 0.218
(a)
What is the probability that a student gets no answers correct?
Sol
 20 
P( X  0)   0.250 (0.75) 20 = 0.003171
0
(b)
What is the probability that a student fails the exam (gets less than 40%)
Sol
P(X<=7) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=7)
= 0.003 + 0.021 + 0.067 + 0.134 + 0.19 + 0.202 + 0.169 + 0.112 = 0.898
(c)
What is the probability that a student gets a first class honour in the exam (gets greater
than or equal 70%)
Sol
P(X>=14) = P(X=14) + P(X=15) + P(X=16) + P(X=17) + P(X=18) + P(X=19) + P(X=20)
= 0.0000257 + 0.00000343 + 0.00000036 + 0.000000028 + 0.0000000016 + 0.000000000055 +
0.000000000001 = 0.0000295
5
The number of telephone calls that arrive at a phone exchange is often modelled as a
Poission random variable. Assume that on a that on average there are 8 calls per hour.
(a)
What is the probability that there are exactly 5 calls in one hour?
Sol
(b)
Sol
(c)
Sol
(d)
Sol
P( X  5) 
e 8 85
 0.091603662
5!
What is the probability that there are 3 or fewer calls in an hour?
P( X  3)  e 8 
e 8 101 e 8 10 2 e 8 10 3


 0.04238
1!
2!
3!
What is the probability that there are exactly 15 calls in two hours?
 = 16. P(Y  15) 
e 16 2015
 0.09922
15!
What is the probability that there are exactly 5 calls in 30 minutes?
 = 4. P(Y  5) 
e 4 4 5
 0.15629
5!
6
Astronomers treat the number of stars in a given volume of space as a Poission random
variable. The density in the Milky Way Galaxy in the vicinity of our solar system in one
star per 16 cubic light years.
(a)
What is the probability of two or more stars in 16 cubic light years?
Sol
 = 1. P(X≥2) = 1 – (P(X=0) + P(X=1)) = 0.264
(b)
How many cubic light-years of space must be studied so that the probability of one or
more stars exceeds 0.95?
Sol
In order that P(X≥1) = 1 - P(X=0) = 1 - e- µ exceed 0.95, we need µ=3.
Therefore 3*16=48 cubic light years of space must be studied.
7
Assume X is a normally distributed with a mean of 5 and a standard deviation of 4.
Determine the following:
(a)
P(X<11)


Sol
P(X < 11) = P Z 
11  5 

4 
= P(Z < 1.5)
= 0.93319
(b)
P(X>0)


Sol
P(X > 0) = P Z 
05

4 
= P(Z > 1.25)
= 1  P(Z < 1.25)
= 0.89435
(c)
Sol
P(3<X<7)
7 5
35
Z

4 
 4
P(3 < X < 7) = P
= P(0.5 < Z < 0.5)
= P(Z < 0.5)  P(Z < 0.5)
= 0.38292
(d)
Sol
P(-2<X<9)
P(2 < X < 9)
95
25
Z

4 
 4
= P
= P(1.75 < Z < 1)
= P(Z < 1)  P(Z < 1.75)]
= 0.80128
8
The time until recharge for a battery in a laptop computer under common conditions is
normally distributed with a mean of 260 minutes and a standard deviation of 50 minutes.
(a)
What is the probability that a battery lasts more than four hours?
Sol
X ≈ N(260, 502)
P( X  240)  1  P( X  240)  1  (
240  260
)  1  (0.4)  1  0.3446=0.6554
50
(b)
What are the quartiles (the 25% and 75% values) of the battery life?
Sol
(c)
1 (0.25)  50  260  226.2755
1 (0.75)  50  260  293.7245
What value of life in minutes is exceeded with 95% probability?
Sol
1 (0.05)  50  260  177.7550
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