Probability of Odd Number - Discrete Mathematics and its Applications - Lecture Slides, Slides of Discrete Mathematics

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Lecture 8

5.2 Discrete Probability

5.2 Recap

• Sample space: space of all possible outcomes.
• Event: subset of of S.
• p(s) : probability of element s of S:

• Probability of complement
• Disjoint events:
• Overlapping events:
• Three doors problem (see ex. 39 p. 362 + matlab demo)

∀ s 0 ≤ p s ( ) ≤ 1 ( ) 1

s S

p s

∈

( ) ( ) s E

p E p s ∈

p E ( ) = 1 − p E ( )

1 1

n n i i i i

p E p E

=^ =

 =^ ∑

p E ( 1  E 2 ) = p E ( 1 ) + p E ( 2 ) − p E ( 1  E 2 )

HHH

HHT

HTH

HTT

THH

TTH

THT

TTT

E

F

S

( ) 1 ( ) 1 ( ) 3 ( )^1 p E = 2 P F = 2 P E  F = 4 P E  F = 4

What is the probability of odd number of tails (i.e. of event E) if we know that the first flip was tails (i.e. F happened)?

 If we know that F happened the total number of possible outcomes shrinks to |F| E = THH, THT, TTH, TTT. Of those there are two that make event E happen: THH, TTT. P(E|F) = 1/

THH

TTT

THT

TTH

F

E|F

( ) ( | ) ( )

p E F P E F P F

=



Theorem: If E and F are two events and p(F)>0, then the conditional probability of E given F is given by:

Example: What is the probability that a family with 2 kids has two boys, given that they have at least one boy? (all possibilities are equally likely).

S: all possibilities: {BB, GB, BG, GG}. E: family has two boys: {BB}. F: family has at least one boy: {BB, GB, BG}. E F = {BB}

p(E|F) = (1/4) / (3/4) = 1/

 E^ BB

F

GB

BG

GG

( ) ( | ) ( ) ( ) ( ) ( ) ( ) ( )

P E F P E F P E P E P E F P E P F P F

= ⇔ = ⇔ =

 

Equivalent statement: The probability of the intersection od 2 events is the product of the probabilities of the separate events.

Example: A family has 2 children (|S|=4). Is the event E that the family has children of both sexes independent from the event F that they have at most one boy?

E: {BG, GB} F: {GG, BG, GB}

E (^)  F: {BG, GB}

( ) 1/ 2 ( ) ( ) 1/ 2 3 / 4 3 / 8

P E F P E P F

= = × =



dependent

What about a family with 3 kids?

|S| = 8 |E| = 8 – 2 = 6 (only BBB and GGG violate E). |F| =| {GGG, GBB, BGB, BBG} | = 4

| E 1  E 2 |=| {GBB, BGB, BBG}| = 3

( ) 3 / 8 ( ) ( ) 3 / 4 1/ 2 3 / 8

P E F P E P F

= = × =



independent! (again: it’s hard to get any intuition for this).

Does it sum to 1?

Recall:

! ( ) !( )!

p q n^ n p qk^ n^ k k n k

• = − −

Use q = 1-p to prove the result.

5.2 Random Variables

Many problems deal with real numbers rather than set memberships. E.g. What is the sum of the outcomes of 2 dice? How many times do we expect to 7 1 in bit-strings of length 12?

We were able to answer those questions before, but we now introduce some formal machinery:

Definition: A random variable X is a function (not a variable!) from sample space to the real numbers. It assigns a real number to each possible outcome. (and is not random!).

Example: S={apple, pear, banana}  X(apple) = 1, X(pear) = 2, X(banana) = 3.

X: the number of times heads comes up when we toss a coin 2 times: X(TT) = 0; X(HT) = 1; X(TH) = 1; X(HH) = 2;

Definition: The expected value of a random variable X(s) on the sample space S is given by: ( ) ( ) ( ) s S

E X X s p s ∈

= (^) ∑

Sometimes, it’s more efficient to compute expectations by clumping together all elements of S that result in the same value for X(s).

, ( )

s S X s r

p X r p s

K r p s

∈ =

∑

number of elements in S such that X(s)=r

only true when all p(s) s.t. X(s)=r have equal probability.

generally true

Example: X is number of heads in 2 coin tosses. Expected value? E(X) = 0 x P(X=0) + 1 x P(X=1) + 2 x P(X=2) = 0 + ½ + 2x ¼ = 1. We expect on average that we see 1 head.

Example: X is the value of the number that comes up on a die. Expected value? E(X) = 1 x 1/6 + 2 x 1/6 + 3 x 1/6 + 4 x 1/6 + 5 x 1/6 + 6 x 1/6 = 21/6 = 7/2.

(matlab demo2)