Chapter 4: Introduction to Probability

1. If we have a two-step experiment in which we flip a coin and cast a die, how many outcomes are there for the entire experiment? Using H for heads and T for tails, write out the entire sample space

Answer:

Since the experiment has N = 2 steps with n1 = 2 results on the first step and n2 = 6 on the second, there would be (n1)(n2) = (2)(6) = 12 outcomes for the overall experiment. The sample space is:

S = {(H,1),(H,2),(H,3),(H,4),(H,5),(H,6),(T,1),(T,2),(T,3),(T,4),(T,5),(T,6)}

2. In one of the best-known versions of the card game Solitaire, 7 cards are dealt from the deck of 52. How many different hands of 7 cards can be drawn from 52?

Use R to find the answer.

Answer:

Using the Counting Rule for Combinations, and n = 7 from N = 52, there are 133,784,560 unique hands of 7 cards from a deck of 52. That is

                     fig3.jpg

choose(52, 7)

##   [1]   133784560

3. Consider a class of first-year statistics students in which 20 undergraduates are enrolled. What is the probability that two or more students will be found to share the same birthday? As a simplifying assumption, ignore leap years (such as 2016). Use R to find the answer. Hint: recall that the probability of an event A equals one minus the probability of the complement of that event, Ac: i.e., p(A) = 1 − p(Ac).

Answer:

0.4114. The probability that at least two people (out of 20) share the same birthday is one minus the probability that no one shares the same birthday. Let A be the event that at least two people share the same birthday, and Ac be the event that no one (out of 20) shares the same birthday. Then

capture2.jpg

#Comment1. find the probability of event A-complement (probability

#no one shares the same birthday)

probAcomp = ((365)*(364)*(363)*(362)*(361)*(360)*(359)*(358)

*(357)*(356)*(355)*(354)*(353)*(352)*(351)*(350)

    *(349)*(348)*(347)*(346)) / ((365) ^ 20)

#Comment2. find probability of event A (at least 2 people share

#same birthday) 

probA = 1 - probAcomp

#Comment3. the probability of event A 

probA

## [1] 0.4114

#Comment4. a more convenient way to compute probability of event

#A-complement using prod() function

NewprobAcomp = prod(365 : 346) / (365) ^ 20

#Comment5. use NewprobAcomp to find probability of event A

NewprobA = 1 - NewprobAcomp

#Comment6. the probability of event A

NewprobA

##   [1]   0.4114384

4. The following data from a sample of 100 families show the record of college attendance by fathers and their oldest sons: in 22 families, both father and the son attended college; in 31 families, neither father nor son attended college; in 12 families, the father attended college while the son did not; and in 35 families, the son attended college but the father did not.

a. What is the probability a son attended college given that his father attended college?

Organizing this information in a simple table

figure

Define the following events

Sc is event that the son attended college

Sn is event that the son did not attend college

Fc is event that the father attended college

Fn is event that the father did not attend college

fig_4.jpg

Answer:

0.6471. There is almost a 65% chance that a son attended college given that his father did.

b. What is the probability a son attended college given that his father did not attend college?

fig_5.jpg

Answer:

0.5303. There is a 53% chance that a son attended college if his father did not.

c. Is attending college by the son independent of whether his father attended college?

Answer:

Since p(Sc|Fc) = 0.6471>p(Sc) = 0.5700, the events are not independent. In other words, knowing whether the father attended college tells us something about whether the son is likely to have attended.

5. Consider the following facts: (a) in random testing, you test positive for a disease; (b) all people who have the disease test positive for it; (c) in 5% of the cases, this test shows positive even when a person does not have the disease (that is, there is a 5% chance of a false positive); and (d) in the population-at-large, one person in a thousand has the disease. What is the probability that you have the disease? Hint: define the following events: D is event person has the disease; Dc is event person does not have the disease; P is event person tests positive for the disease.

Answer:

0.0196

fig

Thus, perhaps surprisingly, even if you test positive for this disease, there is less than a 2% chance that you actually have it.