GB-Probability Theory Flashcards
You pick a card from the deck and the dealer picks another one without replacement. If you have a larger number, you win. However if your number is smaller or equal to the dealer then the house wins. What is my probability of winning?
–> There are thirteen different cards 2 to Ace —> If I pick 2, my probability of winning is 0 —->If i pick, 3, my probability of winning is 4/51 —-> if i pick 4, my probability of winning is 8/51 —> My probability of winning: 1/13 (0/51 + 4/51 + 8/51 + 48/51) = (4 / (13 x 51)) (0 + 1 + …+ 12) –> 8/17
100 Passanger airline. nth person holds the ticket for the nth seat. 1st person is drunk and he will choose a seat randomly, if chooses xth seat, then the xth person can choose randomly. What is the probability that 100th person will end up on his seat?
- Start with only two people - there is 50% chance of one being in his seat
- Start with three people -
- 1 2 3
- 2 1 3
- 2 3 1
- 3 2 1
- 50% chance of # 3 getting his seat
- Start with 4 people:
- 1 2 3 4
- 2 1 3 4
- 2 3 1 4
- 2 3 4 1
- 2 4 3 1
- 3 2 1 4
- 3 2 4 1
- 4 2 3 1
- Once again 50%
What is the total number of hands in the game of poker?
- (52 5 ) - Combination sign
- 52! / (47!)(5!)
- = 2, 598,960
In combination, order does not matter.
What is the probability of getting four-of-a-kind in Poker? How many number o hands with four of-a-kind?
- First we can choose the value of the four cards - we have thirteen choices.
- Then we can choose the 5th card to be anything from the remaining 48 cards
- Total of 13 x 48 = 624
- Probability = 624/ 2,598,960
How many number of hands can have a full house? What is the porbabilty of full house?
- Full house is Three of a kind, followed by a pair
- (13 choose 1) (4 choose 3) (12 choose 2) (4 choose 2)
- 13 x 4 x 12 x 6 = 3744
- Probability of full house = 3744/ 2,598,960
How many possible outcomes are there for hands with two pairs in Poker? What is the probability/
- (13 choose 2) (4 chose 2)(4 choose 2)(48)
- 123552
- Probabily = 123552 / 2,598,960
There are now 11 pirates. They loot gold coins and store them in a vault. Still being a democratic bunch, they decide that only a majority-of them (>=6) together can open the safe. So they ask a locksmith to put a certain number of locks on the safe. To access the treasure, every lock needs to be opened. Each lock can have multiple keys - but each key can only open one lock. What is the smallest numbers of locks needed?
- Let’s randomly select 5 pirates from the 11-member group; there must be a lock that none of them has the key to.
- In other words - we must have a “special lock” to which none of the 5 selected pirates has a key and the other six pirates all have keys. Such 5-group pirates are randomly selected.
- Minimum number of locks = (11 choose 5)
- = 462 locks
- Each lock has six keys - which are given to unique 6-memeber subgroups. So each pirate must have (462 x 6) / 11 = 252 keys
Given N points drawn randomly on the circumference of a circle, what is the probability that they are all within a semicircle?
- Let’s start at one point and clockwise label the points as 1, 2,..N.
- The probability that remaining N-1 points from 2 to N are in clockwise semicircle starting at point 1 is 1/2^(N-1)
- Since there are N different starting points we can pick. The probability that N points randomly drawn on the circumference of a circle all are withing a semi-circle is
- N/(2^(N-1))
Job applications to 5 firms. Your 3-year old messes up which cover letter goes into which envelope. What is the probabiltiy that all five letters are mailed to the wrong firm?
- Classic example of Inclusion-Exclusion principle
- Ei is the Probability that at least one letter has the correct envelope
P( U Ei) = Sum 1 to 5 (P Ei) + sum P (E1 E2) + (-1)^6(P(E1 E2 E3..E5)
What is the probability of at least one letter has the correct envelope?
(1) Probability of 1 letter with correct envelope: 1/5 , there is (5 choose 1), hence it is 1
(2) Proabability of 2 letters with correcet envelopes: 1/5*1/4, (5 Choose 2), hence 1/2!
Similarilty for 3, 4 and 5. Probability of at least one letter with correct envelope = 1 - 1/2! + 1/3! - 1/4! + 1/5! = 19/30
Answer: probability of all letters are in the wrong envelope = 1 - 19/30 = 11/30
How many people do we need in a class to make the probability that two people have the same birthday more than 1/2?
- For one individual we have 365 possibilities,
- For n individuals we have 365^n possibilities
- We want to find the number of sequences that have no duplications of birthdays
- 365 x 364 x …x (365-n+1) (where n is the number of people in the room).
- (365 x 364 x ….x (365-n+1) / (365^n) < 1/2
- Smallest such number is 23. Since we care about the duplication (people having the same birthday).
Let x be an integer between 1 and 10^12. What is the probability that the cubic of x ends with 11?
- All integers can be expressed as x = a + 10b
- a is the last digit of x
- x^3 = (a+10b)^3 = a^3 + 30a^2b+ 300ab^2 + 1000b^3
- The unit digit of x^3 depends on a^3
- a must be one
- Tenth digit only depends on 30a^2b, since a = 1
- 30b = 10, 3b = 1,
- b must be 7 (7x3 = 21)
- Last two digits of x = 71
- Which has 1% probability for integers between 1 and 10^12
If you have an unfair coin, which may bias toward either heads or tails at an unknown probability, can you generate even odds using this coin?
- For an unfair coin, we can’t generate even odds usin only one toss.
- However we can look at two tosses:
- We can assign HT to winning and TH to losing
- Both have the same probability: P(HT) = P(TH)
Jason throws darts at a dartboard, aiming for the center. The second dart lands farther from the center than the first. If Jason throws a third dart aiming for the center, what is the probability that the third throw is farther from the center than the first? (assume Jason’s skillfulness is constant)
- Since there are 3 throws there are six possible outcomes. A being the best throw, and C being the worst throw
- ABC
- ACB
- BCA
- BAC
- CAB
- CBA
- Given the scenario - eliminate 4,5,6
- For 1,2,3 third throw is worst than the 1st throw
- From ABC, ACB and BCA, there is 2/3 probability that third throw is worst.
- If we have n+1 throws - the probability that (n+1)th throw is not best is
- (n/(n+1))
A manager annouces that the first person in line will get a free movie ticket if his birthday matches someone who has already bought a ticket. What position in the line will give you the largest change of getting the free ticket? (assume 365 days).
- Assume: You choose to be the nth person in line
- To get a free ticket, all (n-1) people need to have different birthday
- And your birthday needs to match one of those (n-1) birthday
- p(n) = p(first n-1 people have different birthday)*p(my birthaday is among those n-1 people)
- p(n) = (365x 364x …x 365-n+2)/(365^n-1) * (n-1)/365
- However we need to strike a balance, if we increase n, we increase the probability of matching our birthday but there is a probability that someone else’s birthday will match before ours does.
- P(n) > P(n-1), P(n) > P(n+1)
- P(n-1) = (365 * 364 * ..*(365-n-3))/365)*((n-2)/365))
- P(n) = (365*364*…*(365-(n-2))/365)*((n-1)/365)
- P(n+1) = (365*364*..(365-(n-2)*(365-(n-1)))/365)*(n/365))
You should be the 20th person.
We throw 3 dice one by one. What is the probabiltiy that we obtain 3 points in strictly increasing order?
- To have 3 points in strictly increasin order, three points must be different
- Conditioned on three different numbers, the probability of strictly increasing order is simply
- 1/3! = 1/6
- P = P(different numbers in all three throws) x p (increasing order|3 different numbers)
- = (1x(5/6)x(4/6))*(1/6) = 5/54
There is a one amoeba in a pond. After every minute the amoeba may die, stay the same, split into two, or split into 3 with equal probability. What is the probability that the amoeba poplation will die out?
- Let P(E) be the probability that the amoeba population will die out
- P(E) = P(E|F1)*P(F1)+P(E|F2)*P(F2)+P(E|F3)*P(F3)+P(E|F4)*P(F4)
- F1 = dies, F2 = stays the same, F3 = doubles, F4 = Triples
- P(E|F1) = 1 (amoeba actually dies out)
- P(E|F2) = P(E) state does not change
- F3 - amoeba population dies only if both amoeba dies, probability of that happens is P(E)^2
- Similarily P(F4) = P(E)^3
- P(E) = 1*1/4 + 1/4*P(E) + 1/4*P(E)^2 + 1/4*P(E)^3
- 0
- (x-1)(x^2 +2x -1) = 0
- X = sqrt(2) - 1 = .414
*
You are taking out candies one by one from a jar that has 10 red candies, 20 blue and 30 green. What is the probability that there are at least one blue and one green candy left in the jar when you have taken out all the red candies?
- Let Tr, Tb, Tg be the number that the last red, blue and green candies are taken out respectively.
- To have at least 1 blue and 1 green candy left hen all red candies are taken out, we want to have
- Tr < Tb and Tr < Tg
- Pr (Tr < Tb and Tr < Tg)
- These are two mutually exclusive events and we need to satisfy:
- Pr(Tr < Tb < Tg) + Pr(Tr<tg>
</tg><li>Tr < Tb < Tg means that the last candy is green. Since there are sixty candies and each candy is equally likely - we have Pr(Tg = 60) = 30/60 and Pr (Tb = 30) = 20/30</li><li>Similarly, for Tr < Tg<tb>
<li>Pr (Tb = 60) = 20/60 and Pr(Tg = 40) = 30/40</li>
</tb>
</li>
</li></tg>
* (3/6)(2/3) + (2/6)(3/4) = 7/12
Two players A and B alternatively toss a fair coin (A tosses the coin first, then B tosses the coin, then A and then B) The sequence of heads and tails is reacorded. If there is a head followed by a tail (HT subsequence) the game ends and the person who tosses the tail wins. What is the probability that A wins the game?
- A can never win the game in the first toss while B has 1/4 probability of winning the game in first toss.
- Let P(A) be the probabilty that A wins then B wins is P(B) = 1 - P(A)
- P(A) = .5 P(A|H) + .5 P(A|T)
- If A tosses tails then B becomes the first to start the HT sequence - hence we can replace that P(B) = 1- P(A) = P(A|T)
- P(A|H) = .5*0 + .5*(1-P(A|H))
- P(A|H) = 1/3
- P(A) = .5 P(A|H) + .5(1-P(A))
- P(A) = .5*(1/3) + .5(1-P(A))
- P(A) = 4/9 which is less than 1/2 since the A can never win in the first toss.
52 cards are randomly distributed to 4 players with each player getting 13 cards. What is the probabiltiy that eacy of them will have an ace?
- Let’s begin with any one of the four aces; it has probabilty of 52/52 = 1 of belonging to a pile.
- 2nd ace can be any of the remaining 51 cards - among which 39 of them belong to a different than the 1st pile
- 3rd ace can be any of the remaining 50 cards - among which 38 of them belong a different than 1s and 2nd pile
- 4th ace is in a pile different from the first three aces is 13/49
- So the probability that each pile has an ace is:
- 52/52 x 39/51 x 26/50 x 13/49
A gamber starts with an initual fortune of i dollars. On each successive game, the gambler wins $1 with probability p or looses $1 with probability q = 1 - p. He will stop if he either accumulates N dollars or losses all his money. What is the probability that he will end up with N dollars?
- Let Pi be the probabiltiy that the gambler’s fortune will reach N instead of 0.
- The next state is either i+1 with probability p or i-1 with probability q.
- Pi = pPi+1 + qPi-1
- Pi+1 - Pi = (q/p)(Pi - Pi-1) = (q/p)^2 (Pi-1 - Pi-2)
- (q/p)^(i) (P1 - P0)
- We have a bounddry problem P0 = 0 and PN = 1
- P1 = pP2 + qP0 = P2 = (1/p)*(P1) = [1 + q/p] P1
- P3 = [1 + q/p + (q/p)^2] P1
- PN = 1 = [1 + q/p + ..+ (q/p)^(N-1))]
A basketball player is taking 100 free throws. 1 score if the basketball goes in and 0 if it does not. For each of the following throw the probability of her scoring is the fraction of throws she has made so far. Ex. If she has scored 23 points after 40th throw - the probability that she will score in the 41th throw is 23/40. After 100 throws - what is the probabiltiy that she scores exactly 50 baskets?
- Let (n,k) , 1 < k < n be the event that the player scores k baskets after n throws.
- Pnk = P((n,k))
- The solution is simple if we use an induction approach starting n = 3.
- The third throw has probabilty of 1/2 of scoring.
- P(3,1) = 1/2 and P(3,2) = 1/2
- P4,1 = 1/3, P4,2 = 1/3, P4,3 = 1/3
- P5,1 = 1/4, P5,2 = 1/4, P5,3 = 1/4..
- The results indicate that Pnk = 1/(n-1)
- P100,50 = 1/99
If the probability of observing at least one car on a highway during any 20-minute interval is 609/625, then what is the probabilty of observing at least one car during any 5-minute interval? Assume that the probability of seeing a car at any moment is uniform for the entire 20 minutes.
- We can break down 20-minute interval into a sequence of 4 overlapping 5-minute intervals.
- Since observing a car has constant probability, the probability of observing a car in any 5-minute interval is constant.
- Let’s denote that probability p
- Then the probabilty that we do not observe any car in 5-minute interval is 1-p
- The probabilty that we do not observe any car in all four of such independent 5-minutes intervals is
- (1-p)^4 = 1 - 609/625 = 16/625
- Hence p = 3/5 (observering a car in any 5-minute interval).
Two bankers each arrive at the station at some random time between 5am and 6am (arrival time is uniformly distributed). They stay exactly five minutes and then leave. What is the probability they will meet on any given day?
- Draw a square and plot the shaded region when both bankers will have likelihood of meeting.
- It should be 12 diagonal box and 22 half (11 full) boxes.
- Hence the probabilty of them meeting is 23/144
A stick is cut twice randomly, what is the probability that the 3 segments can form a triangle?
- Let’s assume that the stick’s length is 1
- First cut point is x and 2nd cut point is y
- If x < y, then three segments are
- x, y-x, and 1-y
- The conditions to form a triangle are
- x + (y-x) > 1-y, y > 1/2
- x + (1-y) > y -x, y < .5 + x
- y-x + 1-y > x, x < 1/2
- Graph it.
- Probability to form a triangle is 1/4
