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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 smokedand 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