Answers to extra probability problems
i) In order for the first 5 to be on the fourth roll, the previous three rolls need to be something other than fives.
Pr(first 5 on fourth roll) = Pr(not 5 on first, not 5 on second, not 5 on third, 5 on fourth).
Since all rolls are independent, we can break this joint probability into a product of the marginal probabilities.
Pr(not 5 on first, not 5 on second, not 5 on
third, and 5 on fourth)
= Pr(not 5 on first)*Pr(not 5 on second)*Pr(not 5
on third)*Pr(5 on fourth) = (5/6)(5/6)(5/6)(1/6).
ii) We want Pr(sum =10 | sum > 8) = Pr(sum = 10 and sum > 8) / Pr(sum > 8)
Since a sum of 10 is gerater than 8, the event (sum=10 and sum>8) is equivalent to the event (sum=10). Hence, the above probability simplifies to:
Pr(sum=10 | sum>8) = Pr(sum=10) / Pr(sum>8)
Using the probability calculations described in part a of the next problem, we get Pr(sum=10) = 3/36.
The possible sums greater than 8 include 9, 10, 11, and 12. Adding up the probabilities associated with each of these outcomes, we get Pr(sum > 8) = 10/36.
Hence, Pr(sum=10 | sum > 8 ) = (3/36)/(10/36) = 3/10. Notice that this is much larger than 3/36. The extra information that the sum exceeds 8 increases the chance that the sum is 10.
2. Runs of coins
i) Assuming the coin is fair (i.e., Pr (Heads) = Pr(Tails) = 0.5), and assuming that flips are independent, both sequences are equally likely to occur. This is because
Pr(HTHHTHTTTH) = Pr(H) * Pr(T) * Pr(H) * Pr(H) * Pr(T) * Pr(H) * Pr(T) * Pr(T) * Pr(T) * Pr(H) = 0.5 * 0.5 * 0.5 * 0.5 * 0.5 *0.5 * 0.5 * 0.5 * 0.5 * 0.5
Pr(HHHHHHHTTT) = Pr(H) * Pr(H) * Pr(H) * Pr(H) * Pr(H) * Pr(H) * Pr(H) * Pr(T) * Pr(T) * Pr(T) = 0.5 * 0.5 * 0.5 * 0.5 * 0.5 *0.5 * 0.5 * 0.5 * 0.5 * 0.5
ii) Define the events
A = heads appears at least one time in ten flips
B = heads appears zero times in ten flips.
Because events A and B account for all possible outcomes of 10 flips
of the coin,
Pr (A) = 1-Pr(B)
= 1 - Pr (TTTTTTTTTT)
= 1 - (0.5 * 0.5 * 0.5 * 0.5 * 0.5 *0.5 * 0.5 * 0.5 *
0.5 * 0.5)
3. Blood and Marriage
Define the events:
MO = the man has blood type O. (These are capital Os,
not zeros.)
WO = the woman has blood type O.
Assuming that the man's blood type is independent of the woman's blood type, then the probability that both have blood type O is:
P( MO and WO) = P(MO) * P(WO) = .46 * .46 = 0.2116.
Similarly, define the events:
MA = the man has blood type A.
WA = the woman has blood type A.
MB = the man has blood type B.
WB = the woman has blood type B.
MAB = the man has blood type AB.
WAB = the woman has blood type AB.
Then, the ways for both to have the same blood type include: (MO and WO), (MA and WA), (MB and WB), (MAB and WAB)
Assume that the man's blood type is independent of the woman's blood type. We then can compute the probabilities of these events like we did for (MO and WO), then add them to get:
Pr (both same blood type) = Pr (MO and WO) + Pr
(MA and WA) + Pr (MB and WB) + Pr (MAB and WAB)
= .46 * .46 + .40 * .40
+ .10 * .10 + .04 * .04
= 0.3832.
These assumptions are reasonable provided that people do not look for mates that have similar blood types to them. For example, if people look for someone of the same race or ethnicity, and if blood types are in different proportions for different ethnicities (which they are), then the man's blood type will NOT be independent of the woman's blood type. For example, assuming men with blood type AB look for women that also have AB, then knowing that a man has blood type AB means it is more likely that the woman he is with also will have blood type AB.
Moral of the story: assumptions of independence need to be checked thoroughly before they are accepted.
4. Blackjack
There are 52 cards in a deck, so that the total number of outcomes for the first card is 52. Each is equally likely to be picked.
(i) There is only one way to get a Jack and a heart: get the Jack of hearts. Hence, Pr (Jack and heart) = 1/52.
(ii) There are 16 ways to get a Jack or a hearts: get one of the thirteen hearts (Ace through King of hearts), or get one of the Jack of clubs, Jack of spades, or Jack of diamonds. Hence, Pr(Jack or hearts) = 16/52. Notice that the Jack of hearts is counted in the thirteen hearts cards, so that I didn't count it again when listing the Jacks.
(iii) There are thirteen hearts. There is only one Jack among those 13 hearts: the Jack of hearts. If we know the first card is a hearts, then the chance that it is a Jack is the number of ways to get the Jack of hearts out of the total number of hearts. Hence, Pr(Jack | hearts) = 1/13.
(iv) There are four Jacks. There is only one heart among these four Jacks: the Jack of hearts. If we know the first card is a Jack, then the chance that it is a heart is the number of ways to get the Jack of hearts out of the total number of Jacks. Hence, Pr(Jack | hearts) = 1/4.
(v) To win, you have to get a 21. There are several ways to get 21: draw a three in one card, draw a two and an Ace in two cards, and draw three Aces in three cards.
Getting 21 by drawing a three:
Since there are four threes in the deck, and 48 cards remaining after the first four cards are dealt, the chance of getting a 3 is Pr(get a 3) = 4/48.
Getting 21 by drawing an Ace first and a 2 second:
Pr(get Ace first and a 2 second) = Pr (get a 2 second | get an Ace
first) * Pr (get an Ace first)
= (4 / 47) * (4/48)
The 4/48 is determined as follows. Since there are four aces in the deck, and 48 cards remaining after the first four cards are dealt, the chance of getting an Ace first is 4/48.
The 4/47 is determined as follows. Since there are four 2s in the deck, and 47 cards remaining after the first five cards are dealt, the chance of getting a 2 on the next card is 4/47.
Getting 21 by drawing a 2 first and an Ace second.
By similar logic,
Pr(get 2 first and Ace second) = Pr (get Ace second | get 2
first) * Pr (get 2 first)
= (4 / 47) * (4/48)
Getting 21 by drawing three Aces.
There are 4*3*2 = 24 ways to get three aces, and there are (48*47*46) ways to pick three cards. Hence, the probability of picking three aces is 24/(48*47*46).
Hence, Pr(win) = 4/48 + (4*4)/(47*48) +
(4*4)/(47*48) + (4*3*2)/(47*48*46) = .0907.
5. Sex of Children
i) Because the babies' sexes are independent, Pr ( M, F, M, F, M, F) = Pr(M) * Pr(F) * Pr(M) * Pr(F) * Pr(M) * Pr(F) = .4986 * .5014 * .4986 * .5014 * .4986 * .5014
ii) Because the babies' sexes are independent, Pr ( F, F, F, F, F, F) = Pr(F) * Pr(F) * Pr(F) * Pr(F) * Pr(F) * Pr(F) = .4986 * .4986 * .4986 * .4986 * .4986 * .4986
iii) Because the babies' sexes are independent, Pr ( F | M, M,
M ) = Pr (F) = .5014
6. For the future lawyers
The prosecutor's probability calculation is correct only if the events wearing Duke sweatpants and Carolina sweatshirt are independent. This is not likely to be true in Chapel Hill!!! People are either Carolina fans or Duke fans. It is likely that Pr(wear Carolina sweatshirt | wear Duke sweatpants) is pretty small, so that the joint probability of wearing both is pretty small. Hence, the defense's argument for reasonable doubt cannot be based on clothing.
When two events A and B are not independent, it is not true that Pr (A and B) = Pr (A) * Pr(B).
7. A Lear jet, a mansion, and a big, big, big pool.
There are a total of 1000 numbers going from 000 to 999. Since each digit (0 to 9) is equally likely to occur for all three parts of the number (the 100s place, 10s place, and 1s place), each number must be equally likely to occur.
Hence, the chance of picking any particular number is 1/1000.
A more formulaic way to calculate this is in two steps:
Step 1:
Pr(first digit is a 1 and second digit is a 2) = Pr(first digit is a 1) * Pr(second digit is a 2 | first digit is a 1) = 1/10 * 1/10 = 1/100.
(Note: what we get on the second digit actually does not depend on
what we got on the first digit.)
Step 2:
Pr(third digit is a 3 and second digit is a 2 and first digit is a
1) =
Pr(third digit is a 3 | second digit is a 2 and first digit is a 1) *
Pr(second digit is a 2 and first digit is a 1) = 1/10 * (1/100) =
1/1000.
i) Straight style: win only when 123 drawn. Pr(win) = 1/1000.
Box style: win when any of 123, 132, 213, 231, 312, 321 drawn. Pr(win) = 6/1000.
Straight/Box style: win
when any of 123, 132, 213, 231, 312, 321 drawn.
Pr(win) = 6/1000.
ii) Straight style: win only when 122 drawn. Pr(win) = 1/1000.
Box style: win when any of 122, 221, 212 drawn. Pr(win) = 3/1000.
Straight/Box style: win
when any of 122, 221, 212 drawn. Pr(win)
= 3/1000.
iii) Straight style: win only when 222 drawn. Pr(win) = 1/1000.
Box style: win when 222 drawn. Pr(win) = 1/1000.
Straight/Box style: win when 222 drawn. Pr(win) = 1/1000.
These problems demonstrates independence, which we learn about on
Thursday. Two variables are independent when the probability that
one occurs is not affected by by whether the other event occurs
(e.g., the chance that we get a 2 on the second digit is not affected
by what we got on the first digit).
8. The Hardy-Weinberg Law of Genetics
Let the event C = offspring gets AA. For C to happen, the child must get an A from both parents. Let's write all the ways that this can happen:
(child is AA and father is AA and mother is AA)
(child is AA and father is AA and mother is Aa)
(child is AA and father is AA and mother is aA)
(child is AA and father is Aa and mother is Aa)
(child is AA and father is Aa and mother is aA)
(child is AA and father is Aa and mother is AA)
(child is AA and father is aA and mother is Aa)
(child is AA and father is aA and mother is aA)
(child is AA and father is aA and mother is AA)
The child cannot be AA when either the mother or father is aa.
So, Pr(C) = Pr(child is AA and father is AA and mother is AA)
+ Pr (child is AA and father is AA and mother is Aa)
+ Pr (child is AA and father is AA and mother is aA)
+ Pr(child is AA and father is Aa and mother is Aa)
+ Pr(child is AA and father is Aa and mother is aA)
+ Pr(child is AA and father is Aa and mother is AA)
+ Pr(child is AA and father is aA and mother is Aa)
+ Pr(child is AA and father is aA and mother is aA)
+ Pr(child is AA and father is aA and mother is AA)
Let's consider the last probability, Pr(child is AA and father is aA and mother is AA). Using the rule we discussed in class, Pr (X and Y) = Pr(Y|X) * Pr(X), we can restate this joint probability as:
Pr(child is AA and father is aA and mother is AA) = Pr(child is AA | father is aA and mother is AA) * Pr( father is aA and mother is AA)
We can write similar statements for each joint probability in the equation for Pr(C). Assuming that genes are inherited independently, then the chance of getting an A from the father is independent of the chance of getting an A from the mother. Hence,
Pr(AA | father is aA and mother is AA) = Pr( get an A from father and an A from mother | father is aA and mother is AA)
= Pr( get an A from father | father is aA ) * Pr( get an
A from mother | mother is AA)
= 0.5 * 1
since gene A has a 50% chance of being inherited from the father's gene, aA.
Finally, for each father/mother pair, we can compute the Pr(C | father/mother pair)
This answers part 1 of the question.
For part 2 of the question, we need to multiply each of the conditional probabilities above by the chance of getting the corresponding father/mother pair.
Assuming independence of mating, then
Pr(father is aA and mother is AA) = Pr(father is aA) * Pr (mother is AA) = q0 * p0.
Similar computations can be used to derive that:
Father Mother
-----------
AA AA
Pr(AA, AA) = p0 * p0.
AA Aa
Pr(AA, Aa) = p0 * q0.
AA aA
Pr(AA, aA) = p0 * q0.
Aa
AA Pr(Aa, AA) = q0
* p0.
Aa
Aa Pr(Aa, Aa)
= q0 * q0.
Aa
aA Pr(Aa, aA)
= q0 * q0.
aA
AA Pr(aA, AA) = q0
* p0.
aA
Aa Pr(aA, Aa)
= q0 * q0.
aA
aA Pr(aA, aA)
= q0 * q0.
Adding up the products of the part 1 and part 2, we get:
Father
Mother
-----------
AA AA
Pr(C | AA, AA) * Pr(AA, AA) = 1 * p0 * p0.
AA Aa
Pr (C | AA, Aa) * Pr(AA, Aa) = 0.50 * p0 * q0.
AA aA
Pr (C | AA, aA) * Pr(AA, aA) = 0.50 * p0 * q0.
Aa
AA Pr (C |
Aa, AA) * Pr(Aa, AA) = 0.50 * q0 * p0.
Aa
Aa Pr (C | Aa, Aa)
* Pr(Aa, Aa) = 0.25 * q0 * q0.
Aa
aA Pr(C | Aa,
aA) * Pr(Aa, aA) = 0.25 * q0
* q0.
aA
AA Pr(C |
Aa, aA) * Pr(aA, AA) = 0.50
* q0 * p0.
aA
Aa Pr(C | aA,
Aa) * Pr(aA, Aa) = 0.25
* q0 * q0.
aA
aA Pr(C | aA,
aA) * Pr(aA, aA) = 0.25
* q0 * q0.
-------------------- ---------
Pr(C)
= p0 * p0 + 2
* p0 * q0
+ 0.5 * q0 * q0.