Question

Problem 1

Suppose there is a bucket with 1 blue, 1 green, 1 red, and 1 purple ball. I draw two out randomly.

What is the sample space of this experiment?

Let R be the event that one of the balls you selected was red. What is P(R)?

Let B be the event that one of the balls you selected was blue. What is P(B)?

Now find: ("C" in B^C is the complement. The complement of a set is the set of elements in the population that are not in A. It is denoted as A^C. Example: Population: Integers from 1 to 10 A = {1, 2, 9} A^C = {3, 4, 5, 6, 7, 8, 10})

• P ( B^C) =

• P(R ∩ B) =

• P(R ∪ B) =

• P(R | B) =

Based on the values that you have just calculated, are events R and B statistically independent? Explain your answer.

Problem 3

A social worker observed that people in a drug addiction treatment program who exercise regularly are less likely to relapse than those who do not. Support the probability of relapse during the first year after quitting is 0.50, and the probability of relapse among those who exercise regularly is 0.20. You also know that half of the subjects exercise regularly.

Define the two “events” in this case.

Write down the unconditional probabilities of these events. Use proper probability notation.

What is the probability of relapse among those who do not exercise regularly?

Answer 1

Problem 1 -

There are - 1 blue, 1 green, 1red and 1 purple balls.

let us denote the event of selecting 1 blue and 1 green ball as bg and so on.

note that bg is same as gb as order is not important here hence we only include unique outcomes in the sample space.

Sample space when we select 2 balls at random - {  bg, br, bp, gr, gp, rp }

R: Event that 1 of the balls selected is red - { br, gr, rp }

B: Event that 1 of the balls selected is blue - { bg, br, bp }

R^C: Event that red ball is not selected - { bg, bp, gp }

B^C: Event that blue ball is not selected - { gr, gp, rp }

: Event that both red and blue balls are selected - { br }

: Event that either red or blue ball is selected - { bg, br,bp, gr, rp }

P(R) = number of outcomes in event R / total number of outcomes in sample space = 3/6 = 1/2

P(B) = number of outcomes in event B / total number of outcomes in sample space = 3/6 = 1/2

P(B^C) = number of outcomes in event B^C / total number of outcomes in sample space = 3/6 = 1/2

P() = number of outcomes in event / total number of outcomes in sample space = 1/6

p() = number of outcomes in event / total number of outcomes in sample space = 5/6

P(R|B) = P() / P(B) = 1/6 / 1/2 = 2/6 = 1/3

To be statistically independent events P(R|B) should be equal to P(R) which is not the case. Hence R and B are not statistically independent.

This can also be understood as if we know 1 of the balls selected is blue then it influences the likelihood of the event that one ball out of 2 selected is red since our sample space is reduced, hence they are not independent.

Problem 3 -

Events - let us define our 2 events as

1. R: Relapse in drug addiction
2. E: Excercise regularly

Unconditional probabilities -

1. P(R) = 0.5
2. P(E) = 0.5

Probability of relapse among those who do not excercise regularly = P(R|E^C)

We are given P(R|E) = 0.2

Since conditional probabilities follow the laws of probability and P(R|E) and P(R|E^C) form exhaustive set,

P(R|E) + P(R|E^C) = 1

hence, P(R|E^C) = 1 - P(R|E) = 1 - 0.2 = 0.8

We can also check this another way -

P(R|E) = P(R and E) / P(E) = 0.2 hence P(R and E) = 0.2 * P(E) = 0.2 * 0.5 = 0.1

P(R and E^C) = P(R) - P(R and E) = 0.5 - 0.1 = 0.4

Hence P(R|E^C) = P(R and E^C) / P(E^C) = P(R and E^C) / (1 - P(E)) = 0.4 / (1-0.5) = 0.4/0.5 = 0.8