4. Conditional Probability, Schemes and Mind Maps of Probability and Statistics

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4. Conditional Probability

BT 1.3, 1.

CSE 312

Spring 2015

W.L. Ruzzo

P( )

conditional probability - intuition Roll one fair die. What is the probability that the outcome is 5? 1/6 (5 is one of 6 equally likely outcomes) What is the probability that the outcome is 5 given that the outcome is an even number? 0 (5 isn’t even) What is the probability that the outcome is 5 given that the outcome is an odd number? 1/3 (3 odd outcomes are equally likely; 5 is one of them) Formal definitions and derivations below

dice Roll one fair die. What is the probability that the outcome is 5 given that it’s odd? E = {5} event that roll is 5 F = {1, 3, 5} event that roll is odd Way 1 (from counting): P(E | F) = |EF| / |F| = |E| / |F| = 1/ Way 2 (from probabilities): P(E | F) = P(EF) / P(F) = P(E) / P(F) = (1/6) / (1/2) = 1/ Way 3 (from restricted sample space): All outcomes are equally likely. Knowing F occurred doesn’t distort relative likelihoods of outcomes within F, so they remain equally likely. There are only 3 of them, one being E, so P(E | F) = 1/

dice Roll a fair die. What is the probability that the outcome is 5? E = {5} (event that roll is 5) S = {1,2, 3, 4, 5, 6} sample space P(E) = |E| / |S| = 1/ 6 What is the prob. that the outcome is 5 given that it’s even? G = {2, 4, 6} Way 1 (counting): P(E | G) = |EG| / |G| = |∅| / |G| = 0/3 = 0 Way 2 (probabilities): P(E | G) = P(EG) / P(G) = P(∅) / P(G) = (0) / (1/2) = 0 Way 3 (restricted sample space): Outcomes are equally likely. Knowing G occurred doesn’t distort relative likelihoods of outcomes within G; they remain equally likely. There are 3 of them, none being E, so P(E | G) = 0/

slicing up the spam

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

G = user #3 receives 5 spam emails

What is P(G|F)?

= 0

conditional probability - general definition

General defn: where P(F) > 0

Holds even when outcomes are not equally likely.

Example: S = {# of heads in 2 coin flips} = {0, 1, 2}

NOT equally likely outcomes: P(0)=P(2)=1/4, P(1)=1/

Q. What is prob of 2 heads (E) given at least 1 head (F)?

A. P(EF)/P(F) = P(E)/P(F) = (1/4)/(1/4+1/2) = 1/

Same as earlier formulation of this example (of course!)

chain rule example - piling cards

piling cards Deck of 52 cards randomly divided into 4 piles 13 cards per pile Compute P(each pile contains an ace) Solution: E 1 = { in any one pile } E 2 = { & in different piles } E 3 = { in different piles } E 4 = { all four aces in different piles } Compute P(E 1

E

2

E

3

E

4

E

1 = { in any one pile } E 2 = { & in different piles } E 3 = { in different piles } E 4 = { all four aces in different piles } P(E 1 E 2 E 3 E 4 ) = P(E 1 ) P(E 2 |E 1 ) P(E 3 |E 1 E 2 ) P(E 4 |E 1 E 2 E 3 ) P(E 1 ) = 52/52 = 1 (A♥ can go anywhere) P(E 2 |E 1 ) = 39/51 (39 of 51 slots not in A♥ pile) P(E 3 |E 1 E 2 ) = 26/50 (26 not in A♥, A♠ piles) P(E 4 |E 1 E 2 E 3 ) = 13/49 (13 not in A♥, A♠, A♦ piles) piling cards A conceptual trick: what’s randomized? a) randomize cards, deal sequentially into 4 piles b) sort cards, aces first, deal randomly into empty slots among 4 piles.

piling cards E 1 = { in any one pile } E 2 = { & in different piles } E 3 = { in different piles } E 4 = { all four aces in different piles } P(E 1

E

2

E

3

E

4

= P(E

1

) P(E

2

|E

1

) P(E

3

|E

1

E

2

) P(E

4

|E

1

E

2

E

3

sending bit strings

sending bit strings Bit string with m 1’s and n 0’s sent on the network All distinct arrangements of bits equally likely E = first bit received is a 0 F = k of first r bits received are 0’s What’s P(E|F)? Solution 1 (“restricted sample space”): Observe: P(E|F) = P(picking one of k 0’s out of r bits) So: P(E|F) = k/r