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STATISTICS 2R PROBABILITY

Workshop 1 (with Solutions)

Equally-likely outcomes models

Task 1. Computers  store  and  send  data  in  binary  form,  i.e.  as  a sequence of 0’s and 1’s. A binary“word”of length 8 is called a byte (e.g. 01101010).

(a) Find the probability that a randomly-generated byte contains only 0’s.

The sample space contains k = 28 = 256 equally-likely outcomes. Let A be the event‘a byte contains only 0’s’. Then A = {00000000}.

n(A) = 1 P (A ) = = =0.00391

(b) Find the probability that a randomly-generated byte contains exactly three 1’s.

Let B be the event‘a byte contains exactly three 1’s’.

The number of different permutations of 3 1’s and 5 0’s is =56.

n(B) = 56 P ( B ) = = =0.2188 .

(c) Find the probability that a randomly-generated byte contains at most three 1’s,

Let C be the event‘a byte contains at most three 1’s’. Then:

n (C) = + + + =1+8 + 28 +56=93 ;

P (C )== =0.3633 .

(d) Find  the  probability  that  a  randomly-generated  byte  consists  of alternating 0’s and 1’s.

This event is D = {01010101, 10101010}, so P (D )= =0.00781 .

Using the rules of probability

Task 2. Suppose you own two bicycles, a new one and an old one. Suppose that the new one has a flat tyre 5% of the time, the old one has a flat tyre 10% of the time, and both have flat tyres 2% of the time.

(a) What is the probability that at least one of your bicycles has a flat tyre?

Let N =‘new bike has flat tyre’, O =‘old bike has flat tyre’.

Given that P(N) = 0.05, P(O) = 0.10, P(NO) = 0.02.

So P(NO) = P(N) + P(O) – P(NO) = 0.05 + 0.10 – 0.02 = 0.13

(b) What is the probability that neither bicycle has a flat tyre?

P(NO′) = P[(NO)′] = 1 – P(NO) = 1 – 0.13 = 0.87

Task 3. In Central Europe, ticks can carry two diseases that can be transmitted to humans: tick-borne encephalitis (TBE) and Lyme disease (borreliosis).  Both diseases affect the central nervous system and they are incurable unless treated immediately.  Suppose there is probability 0.05 that a tick carries TBE, 0.02 that a tick carries both TBE and Lyme disease,  and  0.67  that  a  tick  carries  neither  disease.   What  is  the probability that a tick carries Lyme disease?

Let T =‘tick carries TBE’, L =‘tick carries Lyme disease’.

Given that P(T) = 0.05, P(T L) = 0.02, P[(T L)′] = 0.67.

So         P(T L) = 1 – P[(T L)′] = 0.33

P(T L) = P(T) + P(L) – P(TL)

P(L) = P(T L) + P(TL) – P(T)

= 0.33 + 0.02 – 0.05 = 0.3

Deriving theoretical results from the Axioms

Task 4. Suppose that E1  and E2  are disjoint events in the sample space

S.  Let E3  = (E1 E2)′ . Working from the Axioms of Probability, prove: P(E1) + P(E2) + P(E3) = 1.

E1 , E2 and E3 are disjoint events.

So P(E1 E2 E3) = P(E1) + P(E2) + P(E3)         [Axiom 3]

But E E2 E3 = (E1 E2) (E1 E2)′ = S

P(EE2 E3) = P(S) = 1 [Axiom 2]

So P(E1) + P(E2) + P(E3) = 1

Task 5. Suppose that E1 , E2  and E3  are any three events in the sample space S. Working from the Axioms of Probability, prove that:

P(E1 E2 E3) = P(E1) + P(E2) + P(E3) – P(E1 仁E2)

– P(E1 仁E3) – P(E2 仁E3) + P(E1 仁E2 仁E3).

[Hint: draw a Venn Diagram and note that E1 E2 E3  can be written as the union of seven disjoint events; this is an alternative proof to the one required in Example 9 of the example sheet.]

E1 UE2 UE3 is the union of the following disjoint events:

E1 OE2 OE3

E1 OE2 OE3

E1 OE2 OE3

E1 OE2 OE3

E1 OE2 OE3

E1 OE2 OE3

E1 OE2 OE3

So P(E1 UE2 UE3) = P(E1 OE2 OE3 ′) + P(E1 OE2 OE3 ′) + P(E1 OE2 OE3) + P(E1 OE2 OE3) + P(E1 OE2 OE3 ′) + P(E1 OE2 OE3)

+ P(E1 OE2 OE3)                                       [Axiom 4]

But   E1 = (E1 OE2 OE3 ′) U(E1 OE2 OE3 ′) U(E1 OE2 OE3) U(E1 OE2 OE3)

P(E1) = P(E1 OE2 OE3 ′) + P(E1 OE2 OE3 ′) + P(E1 OE2 OE3) + P(E1 OE2 OE3) [Axiom 4]

There are similar expressions for P(E2) and P(E3).

Also P(E1 OE2) = P(E1 OE2 OE3 ′) + P(E1 OE2 OE3)

There are similar expressions for P(E1 OE3) and P(E2 OE3).

Substituting  these  expressions  into  the first  equation  and  simplifying gives the required result.