Elementary Statistics – More Probability Problems

 

  1. Students sometimes confuse the ideas of disjointness and independence. Recall two events are disjoint if they share no outcomes in common, and two events are independent if knowing one event occurs does not change the probability the other occurs.

 

    1. Suppose two events, A and B, (each with positive probability) are disjoint (for example, rolling an even number on a 6-sided die and rolling an odd number on a 6-sided die). Determine P(A|B). Are A and B independent events?

 

 

 

    1. At a local college, 52% of students are female, 60% of students are from Wisconsin, and 15% are English majors. Furthermore, 31.2% are female students from Wisconsin, and 10% are female English majors. (You can think of these percentages as probabilities.) Are the events {female} and {Wisconsin native} independent? Are the events {female} and {English major} independent?

 

 

 

    1. Note part a shows that disjoint events must be dependent, and part b shows that non-disjoint events can be either independent or dependent.

 

 

  1. What happens when outcomes in a sample space are not equally likely? Suppose an unfair coin is flipped three consecutive times, and each time the upward face is recorded. For this coin, P(heads) = 0.7 and P(tails) = 0.3.

 

    1. Write out the sample space for this experiment.

 

 

 

    1. Let A = {tail on the first flip} and B = {exactly two tails in the three flips}. Determine P(B|A). (Note this conditional probability is not 0.5, which is what it would be if the sample space outcomes were equally likely. Furthermore, this conditional probability is not 0.189—if you got this incorrect answer, you erroneously assumed A and B were independent when computing P(A and B).)

 

 

 

    1. Are the events A and B (defined in part b) independent? (In this example, the coin flips are independent, but the compound events A and B are not independent.)

 

 

 

  1. A health study tracked a group of people for five years. At the beginning of the study, 35% were classified as smokers and 65% were classified as nonsmokers. Results of the study showed that smokers were twice as likely to die as nonsmokers during the five-year study. Given that a randomly selected participant dies over the five-year period, determine the conditional probability the participant was a smoker. (Hint: Use a tree diagram.)

Elementary Statistics – Solutions to Probability Problems

 

  1. Students sometimes confuse the ideas of disjointness and independence. Recall two events are disjoint if they share no outcomes in common, and two events are independent if knowing one event occurs does not change the probability the other occurs.

 

    1. Suppose two events, A and B, (each with positive probability) are disjoint (for example, rolling an even number on a 6-sided die and rolling an odd number on a 6-sided die). Determine P(A|B). Are A and B independent events?

 

If A and B are disjoint events, then P(A and B) = 0 (since they share no outcomes in common). Hence,, so the events are dependent. That is, if two events are disjoint, then they must be dependent (since knowing one of the events occurs definitely changes the probability of the other occurring).

 

    1. At a local college, 52% of students are female, 60% of students are from Wisconsin, and 15% are English majors. Furthermore, 31.2% are female students from Wisconsin, and 10% are female English majors. (You can think of these percentages as probabilities.) Are the events {female} and {Wisconsin native} independent? Are the events {female} and {English major} independent?

 

Note. This conditional probability is equal to the unconditional probability, P(female) = 0.52. Hence, knowing a student is from Wisconsin does not change the probability the student is female. That is, these events are independent. (Alternatively, we could have calculated P(Wisconsin | female) = (0.312)/(0.52) = 0.6 and compared it to P(Wisconsin) = 0.6. Note we would reach the same conclusion.)

 

 

Now note. This conditional probability is not equal to the unconditional probability, P(female) = 0.52. Hence, knowing a student is an English major makes it more likely the student is female. That is, these events are dependent. (Alternatively, we could have calculated P(English | female) = (0.1)/(0.52) 0.19 and compared it to P(English) = 0.15. Note we would reach the same conclusion.)

 

 

    1. Note part a shows that disjoint events must be dependent, and part b shows that non-disjoint events can be either independent or dependent.

 

 

 

  1. What happens when outcomes in a sample space are not equally likely? Suppose an unfair coin is flipped three consecutive times, and each time the upward face is recorded. For this coin, P(heads) = 0.7 and P(tails) = 0.3.

 

    1. Write out the sample space for this experiment.

 

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

 


    1. Let A = {tail on the first flip} and B = {exactly two tails in the three flips}. Determine P(B|A). (Note this conditional probability is not 0.5, which is what it would be if the sample space outcomes were equally likely. Furthermore, this conditional probability is not 0.189—if you got this incorrect answer, you erroneously assumed A and B were independent when computing P(A and B).)

 

By definition,. We’ll calculate the numerator and denominator separately:

P(B and A) = P(exactly two tails and a tail on the first flip) = P(THT or TTH) = P(THT) + P(TTH) = P(T)P(H)P(T) + P(T)P(T)P(H) = (0.3)(0.7)(0.3) + (0.3)(0.3)(0.7) = 0.126. (Here we used the specific addition and multiplication rules, since the events THT and TTH are disjoint, and the coin flips are independent.)

 

P(A) = P(tail on first flip) = P(THH or THT or TTH or TTT) = P(THH) + P(THT) + P(TTH) + P(TTT) = P(T)P(H)P(H) + P(T)P(H)P(T) + P(T)P(T)P(H) + P(T)P(T)P(T) = (0.3)(0.7)(0.7) + (0.3)(0.7)(0.3) + (0.3)(0.3)(0.7) + (0.3)(0.3)(0.3) = 0.3. (Again, we used the specific addition and multiplication rules.)

 

Then,.

 

    1. Are the events A and B (defined in part b) independent? (In this example, the coin flips are independent, but the compound events A and B are not independent.)

 

Now, P(B) = P(exactly two tails) = P(HTT or THT or TTH) = P(HTT) + P(THT) + P(TTH) = P(H)P(T)P(T) + P(T)P(H)P(T) + P(T)P(T)P(H) = (0.7)(0.3)(0.3) + (0.3)(0.7)(0.3) + (0.3)(0.3)(0.7) = 0.189.

 

Since P(B | A) = 0.42 is not equal to P(B) = 0.189, these events are dependent. That is, knowing a tail occurs on the first flip increases the chance of getting exactly two tails.

 

 

  1. A health study tracked a group of people for five years. At the beginning of the study, 35% were classified as smokers and 65% were classified as nonsmokers. Results of the study showed that smokers were twice as likely to die as nonsmokers during the five-year study. Given that a randomly selected participant dies over the five-year period, determine the conditional probability the participant was a smoker. (Hint: Use a tree diagram.)

 

Let S represent a smoker, NS represent a nonsmoker, D represent dead, and A represent alive. Then we know P(S) = 0.35, P(NS) = 0.65, and P(D | S) = 2P(D | NS). For simplicity’s sake, we’ll represent P(D | NS) by x, and therefore P(D | S) is represented by 2x. Then we have the following tree diagram (for these web-solutions there are no lines in the tree diagram—you can fill them in yourself):

 

  Smoking status           Health status                                   Event                                     Probability                                                                                                                                                             

                                                                D                                             S and D                                  (0.35)(2x)

                                S

                                                                A                                             S and A                                   (0.35)(1-2x)

 

                                                                D                                             NS and D                               (0.65)(x)

                                NS

                                                                A                                             NS and A                               (0.65)(1-x)

 

 

Then,