STATISTICS 2R PROBABILITY Workshop 1
Hello, dear friend, you can consult us at any time if you have any questions, add WeChat: daixieit
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(N′O′) = 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?
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 介 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. |
2022-07-29