i) A fair, six-sided die is tossed. What is the probability the first 5 occurs on the fourth roll?

ii) Suppose two fair, 6-sided dice are tossed. What is the probability that the sum equals 10 given it exceeds 8?

2.  Runs of coins

For fun on Saturday night, you and a friend are going to flip a fair coin 10 times (geek!).  Let H be the event that a flip lands with heads showing, and let T be the event that a flip lands with tails showing.  Because the coin is fair, assume Pr(H) = Pr(T) = 0.5.  Neither of you know how to flip the coin to obtain some desired outcome.

You flip                 HTHHTHTTTH.
Your friend flips    HHHHHHHTTT.

Problems:

i)   Which sequence is more likely to occur?
ii)   What is the probability that you will get at least one heads in ten flips?
 

3.  Blood and Marriage

Blood comes in four types: O, A, B,  and AB.    The percentages of people in the United States with each blood type are shown below.

Blood Type      Percentage
-----------------
O                             46
A                             40
B                             10
AB                            4

Problems:

i)   What is the probability that two people getting married both have blood type O?    What assumption are you making?
ii)   What is the probability that two people getting married both have the same blood type?   What assumptions are you making?
iii)   Do you think that these assumptions are reasonable?

For more information on blood types, see the web page linked below.

Source:    http://www.er4yt.org/Education/Science_A2_Subtype.html
 

4.  Winning at Blackjack

A standard deck of playing cards has 52 cards.  There are four suits (clubs, diamonds, hearts, and spades), each of which has thirteen numbered cards (2, ..., 9, 10, Jack, Queen, King, Ace).

In the game of blackjack, each card is worth an amount of points.  Each numbered card is worth its number (e.g., a 5 is worth 5 points); the Jack, Queen, and King are each worth 10 points; and the Ace is either worth your choice of either 1 point or 11 points.   The object of the game is to have more points in your set of cards than your opponent without going over 21.   Any set of cards that sum greater than 21 automatically loses.

Here's how the game is played.  You and your opponent are each dealt two cards.   Usually the first card for each player is dealt face down, and the second card for each player is dealt face up.   After the initial cards are dealt, the first player has the option of asking for another card or not taking any cards.  The first player can keep asking for more cards until either he or she goes over 21, in which case  the player loses, or stops at some number less than or equal to 21.  When the first player stops at some number less than or equal to 21, the second player then can take more cards until matching or exceeding the first player's number without going over 21, in which case the second player wins, or until going over 21, in which case the first player wins.

We're going to simplify the game a little and assume that all cards are dealt face up, so that all cards are visible.  This is a wimpier game than the face-down one, but it makes for easier probability calculations!

Problems:

In all these questions, assume your opponent is dealt cards and plays first.

i)  What is the chance that the first card will be a heart and a Jack?

ii) What is the chance that the first card will be a heart or a Jack?

iii) Given that the first card is a heart, what is the chance that it will be a Jack?

iv) Given that the first card is a Jack, what is the chance that it will be a heart?

v) Your  opponent is dealt a King and a 10, and you are dealt a Queen and a 8.   Being smart, your opponent does not take any more cards and stays at 20.   What is the chance that you will win if you are allowed to take as many cards as you need?
 

5.  Sex of children

Do certain families have a tendency to have babies of the same sex?    Let's assume that the sexes of babies are independent (e.g., the sex of the first-born baby does not affect the probability the second-born baby is female) , and that the probability that a baby is born female is 0.5014, which is approximately the current percentage.  Assume that this probability is the same regardless of family size (which is an assumption that is not necessarily true).

Problems:

i)  (5 points)   What is the chance that a family with 6 children has them born in alternating sex order, with the oldest being a male?

ii)  (5 points)   What is the chance that a family has 6 girls in a row?

iii)  (5 points)   Given that the first three children are boys, what is the chance that the next child will be a girl?

An article that analyzes the question of sex tendencies is in the winter 2001 issue of the magazine Chance.

6.  For the future lawyers.

In a crime at UNC Chapel Hill, it is determined that the perpetrator is a student who was wearing Duke sweatpants and a Carolina sweatshirt. A student is arrested who was wearing both of these items of clothing on the evening of the crime.

The defense provides evidence that shows the probability that a randomly selected student in Chapel Hill is wearing Duke sweatpants is 1/10, and the probability that a randomly selected student in Chapel Hill is wearing a Carolina sweatshirt is 1/5. The prosecutor concludes that the probability that a student is wearing both Duke sweatpants and a Carolina sweatshirt is (1/10)(1/5) = 1/50 = 2%, which is large enough to cause reasonable doubt for the jury.

Problem:

Do you think the prosecuter's claim about the probability is true or false?  In three sentences or less, justify your conclusions.  Assume the 1/10 and 1/5 are accurate probabilities.

7.  A Lear jet, a mansion, and a big, big, big pool.

I know that lotteries are poor bets from a probabilistic point of view.  But, sometimes when the jackpot gets up there, I play them  anyway.  The fantasies alone are worth the $1 for a ticket.

Let's calculate the probability of winning the California Daily 3 lottery (there is no lottery in NC).   In this lottery, you pick three numbers between 0 and 9.  You can pick the same number multiple times, e.g. you can pick 7-7-5 or 7-7-7.   You choose to play one of three ways:  (i) "straight", which means you win if you pick the three numbers in the exact order that they are drawn, (ii)  "box", which means you win if you pick the three numbers correctly in any order, (iii) "straight/box", which means you win if you pick three numbers that win either the straight or the box styles.  Prizes for each style are determined from the numbers of people who play the game in that style.

For example, say you pick 7-7-5.  If the drawn numbers are 7-7-5, you'd win in all three styles of play.  If the drawn numbers are 7-5-7, you win the box and straight/box style, but do not win the straight style.  If the numbers are 2-0-5, you win nothing.

Problem:

i)   Say you pick 1-2-3.  What is the probability of winning straight style?   What is the probability of winning box style?  What is the probability of winning straight/box style?

ii)   Say you pick 1-2-2.   What is the probability of winning straight style?   What is the probability of winning box style?  What is the probability of winning straight/box style?

iii)   Say you pick 2-2-2.   What is the probability of winning straight style?   What is the probability of winning box style?  What is the probability of winning straight/box style?

Incidently, because each number is equally likely to come up, and you have to share lottery prizes with others who guessed that number, the best picks are numbers that no one else thinks of writing.

Extra Information on Lotteries

The calculations below are more involved and won't be on homeworks, quizzes, or exams.  But, you might find it interesting to see how to compute the probability of winning one of the lotteries with the really big jackpots.

The big kahuna of the California lotteries is the Super Lotto Plus.  (You know it's big because it's not just the Lotto, it's the Super Lotto Plus!)  Here is the description from the lottery web site:

                           "SuperLottoPlus is your chance to win millions of dollars!
                           The jackpot ranges from $7 million to $50 million or more.
                           The jackpot rolls over and grows whenever there is no
                           winner. All you have to do is pick five numbers from 1 to
                           47 and one MEGA number from 1 to 27 and match them
                           to the numbers drawn by the Lottery every Wednesday
                           and Saturday."    Source: http://www.calottery.com/games/superlottoplus/superlottoplus.asp

From this description, you'd think winning was easy, right?  Well, let's calculate the probability of getting the correct combination.  We'll start by finding the probability of picking the 5 numbers from 1 to 47 correctly.

The total number of ways to pick five different numbers between 1 and 47 equals 47*46*45*44*43=184,072,680.   Note that different permutations of the same set of five numbers (e.g., 1-2-3-4-5 and 1-2-3-5-4 and 1-2-4-3-5, etc.) are all counted in this 184 billion.

Unlike in straight Daily 3, in SuperLotto Plus you don't have to guess the exact order.  We have to eliminate duplicate permutations to get the number of distinct combinations of the 5 numbers.    To determine the number of combinations of 5 distinct numbers,  we use the fact that

(number of combinations of 5 distinct numbers) * (number of ways to order each combination of five distinct numbers) = (number of ways to pick five distinct numbers).

You can verify this equation by comparing the number of ways to win with 1-2-3 in the straight and box  Daily 3 calculations.

To save writing, let's label the number of combinations of the 5 numbers as C.   For any of these combinations, there are 5*4*3*2*1 = 120 possible ways to order them.   Thus, the equation translates to:

                  C * 120 = 184,072,680,

so that C = 184,072,680 / 120  = 1,533,939.

Now, not only do we have to draw the five numbers correctly, but we have to guess the MEGA number.  There are 27 possible choices of the MEGA number.  Hence, the total number of winning combinations equals  27 * 1,533,939  =  41,416,353.

Therefore, your probability of winning is 1 / 41,416,353.   It's very, very , very unlikely that  you will win the SuperLotto Plus!!

A next step is to determine whether or not the chance at winning the jackpot is worth the $1 cost of a ticket.  We'll learn how to do this in a few weeks.  Actually, such calculations become pretty complicated when the jackpot can be split among several players.

By the way, check out the lottery site on the frequency of each number coming up.  This is good data for a final project.

8. The Hardy-Weinberg Law of Genetics  

(THIS PROBLEM IS HARDER THAN WHAT WOULD BE ON THE EXAM, BUT IT IS A USEFUL APPLICATION OF PROBABILITY FOR THOSE INTERESTED IN GENETICS)

Each hereditary trait in an offspring depends on a pair of genes, one contributed by the father and the other by the mother.  A gene is either recessive (denoted by a) or dominant (denoted by A).  The hereditary trait is A if one gene in the pair is dominant (AA, Aa, aA), and the trait is a if both genes in the pair are recessive (aa).  Suppose that the probabilities of the father carrying the pairs AA, Aa (which is treated as the same as aA), and aa are p₀  , q₀ , and r₀ , respectively, where p₀ +  q₀ +  r₀ = 1.  The same probabilities hold for the mother.  Also suppose that the gene from each parents is randomly inherited, so that each gene of a pair has a 50%  chance of  being passed on to the offspring.

Assume that matings are random and the genetic contributions of the father and mother are independent.

We're going to show that the corresponding probabilities for a first-generation offspring are:

p₁ =  (p₀+  q₀ / 2)² ;     q₁ =  2(p₀+  q₀ / 2) (r₀+  q₀ / 2) ;      r₁ =  (r₀+  q₀ / 2)²

Problems:

i)  Calculate the probabilities of the offspring getting the gene AA for all sixteen of the possible father/mother pairs, which are shown in the table below.

Father     Mother
-----------
    AA           AA
    AA           Aa
    AA            aA
    AA            aa
    Aa             AA
    Aa             Aa
    Aa             aA
    Aa             aa
    aA             AA
    aA             Aa
    aA             aA
    aA             aa
    aa              AA
    aa              Aa
    aa              aA
    aa              aa

ii)  For each of these pair, calculate the joint probability that the offspring inherits AA and the parents are that pair.

iii)  Sum up all these probabilities to show that  the probability that an offspring gets AA is:  p₁ =  (p₀+  q₀ / 2)² .

The other probabilities can be determined in a similar fashion. See the answers for details.

The Hardy-Weinberg Law is that the above probabilities remain unchanged for all future generations of offspring, i.e.,   p_n = p₁ ,  q_n = q₁ ,  r_n = r₁ for all n>1.   This can be shown by mathematical induction.