STATISTICS 2R: PROBABILITY Workshop 2
Hello, dear friend, you can consult us at any time if you have any questions, add WeChat: daixieit
STATISTICS 2R: PROBABILITY
Workshop 2
Conditional probability and independence
Task 1. Try this task without looking at the lecture notes!
(a) Write down the definition of the conditional probability of an event, A, given another event, B, with P(B) > 0.
P(A | B) = P(A B)/P(B) |
(b) Write down the definition that two events, A and B, are independent.
A and B are independent if P(A B) = P(A) x P(B) |
(c) How can we interpret the independence of two events?
Task 2. A box contains nr red balls and nb blue balls. The red balls
are labelled 1, ... , nr, and the blue balls are labelled 1, ... , nb . A single ball is drawn at random from the box.
(a) Find the probability that the ball is coloured red.
Let E = “ball is red” . Then P(E) = nr /(nr + nb) |
(b) Find the probability that the ball is labelled 1.
Let F = “ball is labelled 1” . Then P(F) = 2 /(nr + nb) |
(c) Find the probability that the ball is labelled 1, given that it is coloured red. [Hint: do not use the formal definition of conditional probability.]
P(F | E) = 1/ nr |
(d) Find the probability that the ball is coloured red, given that it is labelled 1. [Hint: do not use the formal definition of conditional probability.]
P(E | F) = 1/ 2 |
(e) Find conditions on nr and nb such that the events “ball is coloured
red” and “ball is labelled 1” are independent.
For independence, we require:
P(E F) = P(E) x P(F) i.e. 1/(nr + nb) = nr /(nr + nb) x 2 /(nr + nb) i.e. 2nr = nr + nb i.e. nr = nb
So, these events are independent if and only if the box contains an equal number of red and blue balls. |
Task 3. (Exam question 2011) A study carried out on 11 to 15 year-
old schoolchildren in England in 2010 found that 13% of the pupils had recently drunk alcohol and 7% had recently smoked cigarettes. 38% of the pupils who reported drinking alcohol also reported smoking cigarettes. Based on the findings of the study what is the probability that a randomly selected child had neither smoked nor drunk alcohol?
Let S = “child had smoked” and A = “child had drunk alcohol” . We are given that P(S) = 0.07, P(A) = 0.13 and P(S| A) = 0.38. So:
P(child neither smoked nor drank) = P[(SA)′] = 1 – P(SA) = 1 – P(S) – P(A) + P(SA) = 1 – P(S) – P(A) + P(S| A) x P(A) = 1 – 0.07 – 0.13 + (0.38 x 0.13) = 0.8494 |
Task 4. Suppose that A, B and C are events in the sample space, S.
Show that, if P(A| C) > P(B| C) and P(A| C) > P(B| C), then P(A) > P(B).
P(A) = P(AC) + P(AC′) = P(A| C)P(C) + P(A| C′)P(C′) > P(B| C)P(C) + P(B| C′)P(C′) [since P(A| C) > P(B| C), etc.] = P(B) |
Task 5. Suppose that A, B and C are independent events in the
sample space S. Show that (a) the events (AB) and C are independent, (b) the events (AB) and C are independent.
(a) P[(AB)C] = P(ABC) [Associative Law] = P(A)P(B)P(C) [independence of A, B, C] = [P(A)P(B)] x P(C) = P(AB) x P(C) [independence of A, B] :(AB) and C are independent
(b) P[(AB) C] = P[(AC) (BC)] [Distributive Law] = P(AC) + P(BC) – P(ABC) [Proposition 4.4] = P(A)P(C) + P(B)P(C) – P(A)P(B)P(C) = [P(A) + P(B) – P(A)P(B)] x P(C) = P(AB) x P(C) [Proposition 4.4] :(AB) and C are independent |
An application to genetics
Task 6
Human beings each have one of two blood types, Rhesus positive or Rhesus negative. Like many other traits, our blood type is determined by just one gene. This gene appears in 2 forms (or alleles), which we shall denote by R and r. One or other allele of this gene appears on each of an individual's two chromosomes giving 3 genotypes:
• RR (individuals with an R gene on both chromosomes);
• rr (individuals with an r gene on both chromosomes);
• Rr (individuals with one R gene and one r gene, the order of them being unimportant).
Parents each donate one gene to a child. Each parent is equally-likely to donate each gene he/she possesses to a given child (and this process is independent for different children). Parents donate genes to a given child independently of one another.
(a) By drawing a tree diagram, show that the child of an RR and an Rr parent is equally likely to be RR or Rr.
Gene donated by 1st Parent |
Gene donated by 2nd Parent |
|
1 0.5
P(Child is RR) = 0.5 x 1 = 0.5 P(Child is Rr) = 0.5 x 1 = 0.5 |
R r |
Genotypes RR and Rr have the same appearance or phenotype: Rh +ve blood. Only genotype rr has a different appearance: Rh 一ve blood. This effect is called complete dominance, R being the dominant and r the recessive form of the gene. An RR individual is called a pure dominant, rr a pure recessive, and Rr a hybrid.
(b) By drawing a tree diagram, show that two hybrid parents can produce children of any genotype: RR with probability 0.25, Rr with probability 0.5 and rr with probability 0.25.
Gene donated by 1st Parent |
Gene donated by 2nd Parent |
|||
|
R
r |
0.5
0.5
0.5 0.5 |
R r
R
r |
|
P(Child is RR) = 0.5 x 0.5 = 0.25 P(Child is Rr) = (0.5 x 0.5) + (0.5 x 0.5) = 0.5 P(Child is rr) = 0.5 x 0.5 = 0.25 |
2022-07-29