Probability Analyses in Various Scenarios
Explore diverse probability scenarios including chip defect detection, poker hands, card flipping, and birthdays to understand the likelihood of specific events occurring in each case.
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chip defect detection n chips manufactured, one of which is defective k chips randomly selected from n for testing What is Pr(defective chip is in k selected chips) ? |S| = |E| = Pr(defective chip is in k selected chips)
chip defect detection n chips manufactured, one of which is defective k chips randomly selected from n for testing What is Pr(defective chip is in k selected chips) ? Different analysis: Select k chips at random by permuting all n chips and then choosing the first k. Let Ei = event that ith chip is defective. Events E1, E2, , Ek are mutually exclusive Pr(Ei) = 1/n for i=1,2, ,k Thus Pr(defective chip is selected) = Pr(E1) + + Pr(Ek) = k/n.
chip defect detection n chips manufactured, two k chips randomly selected from n for testing What is Pr(a defective chip is in k selected chips) ? two of which are defective |S| = |E| = (1 chip defective) + (2 chips defective) = Pr(defective chip is in k selected chips)
any straight in poker Consider 5 card poker hands. A straight is 5 consecutive rank cards of any suit What is Pr(straight) ? |S| = |E| = Pr(straight) =
card flipping 52 card deck. Cards flippd one at a time. After first ace (of any suit) appears, consider next card Pr(next card = Ace of spades) < Pr(next card = 2 of clubs) ? Maybe Case 1: Take Ace of Spades out of deck Case 1: Take Ace of Spades out of deck Shuffle remaining 51 cards, add Ace of Spades after first ace |S| = 52! (all cards shuffled) |E| = 51! (only 1 place Ace of spades can be added) Case 2: Do the same thing with the 2 of clubs Case 2: Do the same thing with the 2 of clubs |S| and |E| have same size So, Pr(next card = Ace of spades) = Pr(next card = 2 of clubs)
birthdays What is the probability that, of n people, none share the same birthday? |S| = (365)n |E| = (365)(364)(363) (365-n+1) Pr(no matching birthdays) = |S|/|E| = (365)(364)(363) (365-n+1)/(365)n Some values of n n = 23: Pr(no matching birthdays) < 0.5 n = 77: Pr(no matching birthdays) < 1/5000 n = 100: Pr(no matching birthdays) < 1/3,000,000 n = 150: Pr( ) < 1/3,000,000,000,000,000
birthdays What is the probability that, of n people, none share the same birthday as you? |S| = (365)n |E| = (364)n Pr(no birthdays matches yours) = |S|/|E| = (364)n/(365)n Some values of n n = 23: Pr(no matching birthdays) 0.9388 n = 77: Pr(no matching birthdays) 0.8096 n = 253: Pr(no matching birthdays) 0.4995
shell game Should you switch shells? Should you switch shells?
shell game If you don t switch: Probability 1/3 to pick the right shell. |S| = 3, |E| = 1 If you do switch: Probability 2/3 to choose the wrong and then you will switch to the right shell. |S| = 3, |E| = 2 wrong shell initially, YOU SHOULD SWITCH!
conditional probability Conditional probability Conditional probability is the probability that E occurs given that F has already occurred. Conditioning on F Written as P(E|F) Means P(E, given F already observed) Sample space S reduced to those elements consistent with F (i.e. S S F F) Event space E reduced to those elements consistent with F (i.e. E E F F) With equally likely outcomes,
conditional probability General definition: where P(F) > 0. Holds even when outcomes are not equally likely. Implies: Implies: P(EF) = P(E|F) P(F) (chain rule) (chain rule) What if P(F) = 0? P(E|F) undefined (you can t observe the impossible) General definition of Chain Rule Chain Rule:
slicing up the spam 24 emails are sent 6 each to 4 users. 10 of the 24 emails are spam. All possible outcomes equally likely E = user 1 receives 3 spam emails What is P(E) ?
slicing up the spam 24 emails are sent 6 each to 4 users. 10 of the 24 emails are spam. All possible outcomes equally likely E = user 1 receives 3 spam emails F = user 2 receives 6 spam emails What is P(E|F) ?
slicing up the spam 24 emails are sent 6 each to 4 users. 10 of the 24 emails are spam. All possible outcomes equally likely E = user 1 receives 3 spam emails F = user 2 receives 6 spam emails G = user 3 receives 5 spam emails What is P(G|F) ?
sending bit strings Bit string with m 0 s and n 1 s sent on the network All distinct arrangements of bits equally likely E = first bit received is a 1 F = k of first r bits received are 1 s Solution 1:
sending bit strings Bit string with m 0 s and n 1 s sent on the network All distinct arrangements of bits equally likely E = first bit received is a 1 F = k of first r bits received are 1 s Solution 2: Realize P(E|F) = P(picking one of k 1 s out of r bits) P(E|F) = k/r Booyah.
we all have have straights!? Consider 5 card poker hands. There are three players. E = event that you have a straight F = event that opponent #1 has a straight G = event that opponent #2 has a straight What is P(E|FG) ?
