Question

1. (10 marks) (a) Let m events Bi, , Bm form a partition of the sample space Ω and let event A be any event such that A c S2. Then show that given Bi > 0 for j = 1,.. ., m (b) Considcr a clinical trial where group of paticnts arc trcated for depression. As in many such trials a patient has two possible out- comes, in this study a relapse and no relapse. Refer to a relapse as a "failure" and a no relapse as a "success". The trial treats de- pression with Lithium and assumes that each patient in the group has the same probability p of having a success as their outcome. Assume that we do not know the value of p at the start of the clinical trial. Since this value of p is an important source of uncer- tainty in the trial, the clinician conducting the trial partitions the sample space by the possible values of p. She uses the following 11 possibilities 0,1/10,2/10,...,9/10 and 1 for p. Let B; be the event that p for j 1,...,11, and assume that the Bys are all equally probable events. Now let A1 be the event that the first patient in the group has a succesS. Explain how P(A1/Bj) can be expressed as 10 and show that P(Ai) =豆. Note: It may be useful to recall that the sum for a arithmetic series satisfies ri

Answer 1

$P(A)=P(A\cap B_1)+P(A\cap B_2)+P(A\cap B_3)+...+P(A\cap B_m)$

$=\sum_{j=1}^{m}P(A\cap B_j)$

Since $A\cap B_1, A\cap B_2, A\cap B_3...A\cap B_m$ are all mutually exclusive because are mutually exclusive or disjoint So addition theorem applies where $A=(A\cap B_1)U(A\cap B_2)U(A\cap B_3)U...U(A\cap B_m)$

Since A and B_j are dependent we can write

$\sum_{j=1}^{m}P(A\cap B_j) =\sum_{j=1}^{m}P(A/B_j)*P(B_j)$

Hence we have$P(A)=\sum_{j=1}^{m}P(A/B_j)*P(B_j)..............(1)$

b)If B₁ happens then Probability of A₁ ie the probability of success is 0

Similarly if B₂ happens then  Probability of A₁ ie the probability of success is 1/10

Hence in general P(A₁/B_j) ie Probability of A₁ given B_j happens is $\frac{j-1}{10}$

Now to get P(A_1) we apply equation (1)

ie $P(A_1)=\sum_{j=1}^{m}P(A_1/B_j)*P(B_j)$

$=0*1/11+\frac{1}{10}*\frac{1}{11}+\frac{2}{10}*\frac{1}{11}+....+1*\frac{1}{11}$

$=\frac{1}{10}*\frac{1}{11}[1+2+3+...+10]$

$=\frac{1}{10}*\frac{1}{11}[\frac{10*11}{2}]=\frac{1}{2}$ (the formula for sum of arithmetic series from 1 to 10 is used.)