InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
A sample space consists of 9 elementary event `E_1, E_2, E_3 ..... E_8, E_9` whose probabilities are `P(E_1) = P(E_2) = 0. 08` ,`P(E_3) = P(E_4) = 0. 1`, `P(E_6) = P(E_7) = 0. 2` ,`P(E_8) = P(E_9) = 0. 07`. Suppose `A = {E_1,E_5,E_8}`, `B = {E_2, E_5, E_8, E_9}`. Compute `P(A)`, `P(B)` and `P(AnnB)`. Using the addition law of probability, find `P(AuuB)`. List the composition of the event `AuuB`, and calculate, `P(AuuB)` by adding the probabilities of the elementary events. Calculate `P(barB)` from `P(B)`, also calculate `P(barB)` directly from the elementary events of `barB`. |
|
Answer» Here, `P(E_1) = P(E_2) = 0.08` `P(E_3) = P(E_4) = 0.1` `P(E_6) = P(E_7) = 0.2` `P(E_8) = P(E_9) = 0.07` First we will calculate `P(E_5).` As sample space contains events from `E_1` to `E_9.` `:. P(E_1)+P(E_2)+P(E_3)+P(E_4)+P(E_5)+P(E_6)+P(E_7)+P(E_8)+P(E_9) = 1` `=>0.08+0.08+0.1+0.1+P(E_5)+0.2+0.2+0.07+0.07 = 1` `=>0.9+P(E_5) = 1` `=>P(E_5) = 0.1` Now, `P(A) = P(E_1)+P(E_5)+P(E_8) = 0.08+0.1+0.07 = 0.25` `P(B) = P(E_2)+P(E_5)+P(E_8)+P(E_9) = 0.08+0.1+0.07+0.07 = 0.32` `P(AnnB) = P(E_5)+P(E_8) = 0.1+0.07 = 0.17` `:.P(AuuB) = P(A)+P(B)-P(AnnB) = 0.25+0.32-0.17 = 0.4` Now, `AuuB = {E_1,E_2,E_5,E_8,E_9}` `:. P(AuuB) = P(E_1)+P(E_2)+P(E_5)+p(E_8)+P(E_9) = 0.08+0.08+0.1+0.07+0.07 = 0.4` Now, `P(barB) = 1-P(B) = 1-0.32 = 0.68` Also, `barB = {E_1,E_3,E_4,E_6,E_7}` `:.P(barB) = P(E_1)+P(E_3)+P(E_4)+P(E_6)+P(E_7) = 0.08+0.1+0.1+0.2+0.2 = 0.68` |
|