Elementary Statistics – Solutions to Probability and Random Variable Problems

 

  1. P(woman) = 0.52, P(computer science major) = 0.05, P(woman and computer science major) = 0.02

 

    1. P(computer science major|woman) = P(computer science major and woman)/P(woman) .

 

    1. P(woman | computer science major) = P(woman and computer science major)/P(computer science major) = .

 

    1. The events {woman} and {computer science major} are not independent, they are dependent. This is because knowing a person is a woman decreases the chance the person is a computer science major (from 0.05 to 0.04). Alternatively, knowing a person is a computer science major decreases the chance the person is a woman (from 0.52 to 0.40).

 

 

  1. P(S1) = 0.75

P(S2) = 0.25

P(late|S1) = 0.02

P(late|S2) = 0.05

 

Shipping Service                    Package Status                       Event                                      Probability

(unconditional)                        (conditional)       

 

                                                                Late                         (S1 and Late)                          (0.75)(0.02) = 0.015

                    S1

 

                                                                Not Late                 (S1 and Not Late)                  (0.75)(0.98) = 0.735

 

 

                                                                Late                         (S2 and Late)                          (0.25)(0.05) = 0.0125

               

    S2

 

                                                Not Late                 (S2 and Not Late)                  (0.25)(0.95) = 0.2375

 

Then, P(S2|Late) = P(S2 and Late)/P(Late) .

 

           

                                   

  1. S = {(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4), (4,1), (4,2), (4,3), (4,4)}, where each ordered pair represents (Bubba’s roll, player’s roll)

 

Because the die is fair, each of the 16 outcomes is equally likely. There are 6 possible outcomes where the player rolls a higher number than Bubba: (1,2), (1,3), (1,4), (2,3), (2,4), (3,4). Therefore, the probability that Bubba pays the player $8 is .

 

Suppose Bubba charges nothing to play the game and let X be Bubba’s winnings. (Recall, this is the most straight-forward method to solve this type of problem.) Then the probability distribution of X is

Value of X

- 8

0

Probability

 

Then Bubba’s expected/mea profit (with no charge for the game) is    (that is, Bubba pays out $3 per play of the game on average). Since Bubba’s wants his expected profit to be $2 per play of the game, he must charge 3+2 = $5 for each play of the game.

 

 

  1. We need to use the general addition rule: P(American Express or VISA) = P(American Express) + P(VISA) – P(American Express and VISA) = 0.24 + 0.61 – 0.11 = 0.74, or 74%.

 

 

  1. Each day the stock either moves up (U) or down (D), and the movements from day to day are independent.

 

    1. The sample space is S = {(U and U), (U and D), (D and U), (D and D)}. In order for the stock to be at its original price after two days, it would either have to move up, then down; or move down, then up. Hence, P(original price after two days) = P[(U and D) or (D and U)] = P(U and D) + P(D and U) = P(U)P(D) + P(D)P(U) = (0.6)(0.4) + (0.4)(0.6) = 0.48. [Note we can use the specific addition and multiplication rules, because the events (U and D) and (D and U) are disjoint, and the daily stock moves are independent.]

 

    1. The sample space is S = {(U and U and U), (U and U and D), (U and D and U), (U and D and D), (D and U and U), (D and U and D), (D and D and U), (D and D and D)}. In order for the stock to have increased by 1 unit, it would have to go up on two of the days and go down on the other day (and there are three ways this can happen). Hence, P(increase 1 unit in three days) = P[(U and U and D) or (U and D and U) or (D and U and U)] = P(U and U and D) + P(U and D and U) + P(D and U and U) = P(U)P(U)P(D) + P(U)P(D)P(U) + P(D)P(U)P(U) = (0.6)(0.6)(0.4) + (0.6)(0.4)(0.6) + (0.4)(0.6)(0.6) = 0.432.

 

 

  1. 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:

 

  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,  Note that knowing a person died increased the chance they were a smoker (up to 0.52 from 0.35).

 

 

  1. The mean team time is  (Here we used Rule 2 for means.)

 

Recall there aren’t rules for standard deviations, just variances. So in problems like these, we first find the variance, and then take the square root to determine the standard deviation. The variance of the team time is  (Here we use Rule 2 for variances.)

 

Then the standard deviation of the team time is  

 

  1. Let X be the contestant’s winnings. The possible values for X are $0, $100, and $600. Also (since the questions are answered independently),

 

P(X = $0) = P(first question incorrect) = 1 – 0.3 = 0.7

P(X = $100) = P(first question correct and second question incorrect) = (0.3)(1 – 0.05) = 0.285

P(X = $600) = P(first question correct and second question correct) = (0.3)(0.05) = 0.015

(Note these probabilities sum to 1.)

 

Then the probability distribution of X is

 

Value of X

$0

$100

$600

Probability

0.7

0.285

0.015

 

and the expected value of X is