CMSC250

Recall: Expected value of events $a$ with probaility $p$

$E = \sum\limits_{i=1}^{n}p_ia_i$

Take a random anagram of python

If the T and H are next to each other: Win $5

else randomize the remaining 4 letters

If PONY: win $\$20\text{, else lose}$ $10

Discussed last time: lottery

lottery Ticket is $5.

Grand Prize is 1 million

10 second prizes: $10,000

1000 third prizes: $500

10000 fourth prizes: $10

Expected Value with 500,000 people?

Conditional Probaility: Given events $A,B \subseteq S$, the probability of $A$ given $B$ is the probability that $A$ occurs given that we know $B$ occurs

$P(A\vert B) = \frac{P(A\cap B)}{P(B)}$

$P(A\vert B) = \frac{P(A\cap B)}{P(B)}$

$S = \{1,2,3,4,5,6\}, A = \{1,2,5\}, B = \{1,2,3,4\}$

What is $P(A\vert B)$

What is $P(B\vert A)$

$P(A\vert B) = \frac{P(A\cap B)}{P(B)}$

Random Anagram of "PYTHON"

If "_P_____, what is the probability that the 'P' and the 'Y' are next to each other?

Two events $A,B \subseteq S$ are independent if they they do not affect each other.

$P(A\vert B) = P(A)$ and $P(B\vert A) = P(B)$

Disjoint is not the same as independent. Example?

$P(A\vert B) = \frac{P(A\cap B)}{P(B)}$

Suppose a bag has 5 green and 7 purple balls. You take out one and then another. Probability of purple then green?

Consider: for events $A,B$, and outcome in A is either in B or it isn't.

That is $A \cap B$ and $A \cap B^c$ are disjoint

$P(A) = P(A \cap B) + P(A \cap B^c)$

Consider: for events $A,B$, and outcome in A is either in B or it isn't.

$P(A) = P(A \cap B) + P(A \cap B^c)$

$P(A\vert B) = \frac{P(A\cap B)}{P(B)}$

$P(A) = P(A\vert B)P(B) + P(A\vert B^c)P(B^c)$

$P(A) = P(A\vert B)P(B) + P(A\vert B^c)P(B^c)$

Suppose a bag has 5 green and 7 purple balls. You take out one and then another. Probability second is green?