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ECS509U: Probability and Matrices
Week 4 – Probability exercises
Provide solutions in the space given. Use some rough paper to try out solutions before you write
them down in this paper.
Deadline: Wednesday 25/10, 5pm
Name: Jack Thomason
Student number: 220453231
Exercise 1.
Consider a social networking site, such as Facebook. It is known that if a user sends an invitation to add to
their contact list another user, who is unknown to them, there is always a 60% chance that the invitation will
be accepted. From your Facebook account, you send invitations to 4 people that you do not know, to add
them to your contact list.
•
What is the probability that at most one of them will accept?
•
What is the standard deviation in the number of people who will accept your invitation?
(You can assume that an invitation can only be accepted or rejected. You can assume that an invitation being
accepted or rejected does not depend on other invitations being accepted or rejected)
P(acceptance) = 0.6
N=4
i)
We need to find the sum of probabilities of a = 0 and a = 1.
P(x=0) : C(4, 0) • 0.6*0 • 0.4*4 = 0.0256
P(x=1) : C(4, 1) • 0.6*1 • 0.4*3 = 0.1536
P(x<2) = P(x=0) + P(x=1) = 0.0256 + 0.1536 = 0.1792
ii)
For the standard deviation we need to use the formula for binomial distributions, √np(1-p):
N=4
P = 0.6
1-P = 0.4
Standard deviation = √4•0.6•0.4 = √0.96 = 0.9797958971
Exercise 2.
A box contains 8 light bulbs, of which 3 are defective. A bulb is selected at random from the box and tested
and not replaced back in the box. If it is defective, another bulb is selected and tested, until a non-defective
one is chosen. Find the expected number of bulbs chosen.
For this problem we need to determine the expected value for this model.
P(A) = P(choosing non-defective) = 5/8
P(1 - A) = 3/8
The longest this model can run is for 4 selections, as any non faulty selection will result in success, and there
are at most 3 possible delays to this.
a=1
Non-defective - 5/8
a=2
Defective - 3/8
Non-defective - 5/7
a=3
Defective - 3/8
Defective - 2/7
Non-defective - 5/6
a=4
Defective - 3/8
Defective - 2/7
Defective - 1/6
Non-defective - 5/5
In order to find the expected value, we need to use the following formula:
a1•P(x=1) + a2•P(x=2) + a3•P(x=3) + a4•P(x=4) = (1•5/8) + (2•3/8•5/7) + (3•3/8•2/7•5/6) +
(4•3/8•2/7•1/6•5/5)
= 5/8 + 15/28 + 15/56 + 1/14
= 1.5
Exercise 3.
A warehouse stores a large quantity of network cables: 60% of these cables have diameter 8mm, 20% have
diameter 6mm and 20% have diameter 4mm. A sample of 10 cables is picked at random (because of the large
number of cables stored, you can assume that the distribution of cable diameters (60-20-20) remains constant
as the sample is picked). Answer the following:
a) What is the probability that the sample contains 5 cables with diameter 4mm?
b) What is the probability that the sample contains 2 cables or more with diameter 4mm?
P(8mm) = 0.6
P(6mm) = 0.2
P(4mm) = 0.2
N = 10
There is no change to probability as cables are picked.
i)
For this problem we need to find binomial distribution model.
P(A) - probability of picking a 4mm cable
P(A’) - probability of not picking 4mm cable (8mm + 6mm)
P(A) = 0.2
P(A’) = 0.8
P(x = 5) 10C5 • 0.2*5 • 0.8*5 = 0.0264241152
ii)
For this problem I will use a complementary method. I will find the sum of probabilities for 0 and 1 4mm
cables and subtract this from 1.
P(x=0) = 10C0 • 0.2*0 • 0.8*10 = 0.1073741824
P(x=1) = 10C1 • 0.2*1 • 0.8*9 = 0.268435456
P(x< 2) = 0.3758096384
P(x>1) = 0.6241903616
Exercise 4.
A player tosses 3 fair coins. The player wins £6 if 3 heads occur, £4.50 if two heads occur, and £1.50 if only
1 head occurs. For the game to yield an expected profit to the player of £2.50, how much should the player
lose if no heads occur?
We have the expected value of the model and therefore need to fill in the value for not tossing a heads.
L = Loss value
E(X) = £6 • (1/2)*3 + £4.50•(1/2)*3•3 + £1.50•(1/2)*3•3 - £L • (1/2)*3 = £2.50
E(X) = £3/4 + £27/16 + £9/16 + L(1/8) = £2.50
= 3 + L(1/8) = 2.50
= L(1/8) = -0.50 = -4.00
L = -£4.00
Exercise 5.
Edmund Blackadder is blindfolded and shoots four arrows at a target. You can assume that each time he
shoots there is a probability of 0.1 to hit the target and that shots are independent of each other.
a) What is the variance in the number of missed shots? Give the unit of measurement.
b) Every time he hits the target, Blackadder receives 2 gold coins, while every time he misses he needs to
pay 1 gold coin. Compute the probability that after the four shots Blackadder will have profit of more than 2
gold coins.
N=4
P(hit) = 0.1
a)
We need to use the binomial formula for variance = n•p•(1-p)
V(X) = 4 • 0.1 • 0.9 = 0.36
b)
P(X) achieving 2 coin or greater profit
a=0
mmmm
Profit = £-4
P(x = 0) = 1•(0.9*4) = 0.6561
a=1
hmmm - mhmm - mmhm - mmmh (I will use combinations from now on)
Profit = £-1
P(x=1) = 4• (0.1)*1 • (0.9)*3 = 0.2916
a=2
4! / 2!•2! = 6
Profit = £2
P(x=2) = 6• (0.1)*2 • (0.9)*2 = 0.0486
a=3
4! / 3! • 1! = 4
Profit = £4
P(x=3) = 4• (0.1)*3 • (0.9)*1 = 0.0036
a=4
4! / 4! = 1
Profit = £8
P(x=4) = 1• (0.1)*4 • (0.9)*0 = 0.0001
P(X) = 1 - (P(x=0)+P(x=1)) = 0.0523
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95%
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P(X=3)+P(X=4) = 0.0037
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