MidTerm Flashcards
P( A or B ) =
P( A or B ) = P( A ) + P( B ) - P( A and B )
P( A or B or C ) =
P( A or B or C ) = P( A ) + P( Ac and B ) + P( Ac and Bc and C )
| equals what 2 equations?
y - x | > .3
y > x + .3 and y < x - .3
Set A = area(A)
Uni Set = area(Uni)
what is the probability of set A?
area(A)/area(Uni)
when calculating continuous probabilities we use _____ instead of ______
area instead of # of elements
when calculating discrete probabilities we use _____ instead of ______
of elements instead of area
P( A | B ) = ?
P( A | B ) = P( A and B ) / P( B )
P( A & B ) = ?
P( A and B ) = P( B ) * P( A | B )
P( Bc | A ) = ? (if B and Bc are all possibilities)
P( Bc | A ) = 1 - P( B | A )
P( A1 & A2 & A3) = ?
P( A1 and A2 and A3 ) = P(A1) * P( A2 | A1 ) * P( A3 | A1&A2 )
How many decisions (depth) would a tree have with P( A1 & A2 & A3 & A4)
4 deep
Best way to do multiplication rule? When a question asks you for P( A & B & C )
use a decision tree and map out
P( A ) * P( B | A) * P( C | A&B )
br1 * br2 * br3
Total Probability Theorem
P( B ) = ?
Assume A1, A2, A3 …. An are disjoint and are added up to equal the universal set then
P( B ) = P( A1 ) * P( B | A1) +P( A2 ) * P( B | A2 ) + P( A3 ) * P( B | A3 ) ….. + P( An ) * P( B | An)
multiplication rule
P( A1 & A2 & A3 ) = P( A1 ) * P( A2 | A1 ) * P( A3 | A1 & A2 )
total probability rule formula
P( B ) = P( A1 & B ) + P( A2 & B ) + … + P( An & B )
which turns into
P( B ) = P( A1 ) * P( B | A1 ) + P( A2 ) * P( B | A2 ) + … + P( An ) * P( B | An )
Baye’s Law
P( Ai | B ) = [ P( Ai ) * P( B | Ai ) ] / P( B )
and
P( B ) = P( A1 ) * P( B | A1 ) + P( A2 ) * P( B | A2 ) + … + P( An ) * P( B | An )
Given P( B | A ) and we want to find P( A | B ) what should we use?
Baye’s Law
P( Ai | B ) = [ P( Ai ) * P( B | Ai ) ] / P( B )
First step in reading a probability probem
IDENTIFY THE GIVENS
If we know A & B are independent sets:
P( A & B ) = ?
P( A | B ) = ?
P( B | A ) = ?
P( A & B ) = P( A ) * P( B )
P( A | B ) = P ( A )
P( B | A ) = P ( B )
Binomial Probability (if we care about order)
where p is probability of first outcome (Heads) and 1-p is probability of second outcome (Tails) and n is the number of trials and k is the desired number of first outcome desired.
P(k) = p^k * (1-p) ^(n-k)
Binomial Probability (if we don’t care about order)
( n choose k ) p^k (1-p)^(n-k)
(n choose k) = ?
n! / k! (n-k)!
How many combinations can we have with n distinct letters?
n!
How many combinations can we have with n distinct letters and we pick k without replacement?>
n! / ( n-k )!
If we have multiple repeating letters and a total number of letters = t then how do we calc? x y z are the repeating lettters.
t! / #ofx! #ofy! #ofz!
total number of objects = t
total number of objects is partitioned into 3 categories. p1 p2 and p3 (represent the number need for each partition)
how many combinations?
t! / p1! p2! p3!
how to calculate combinations? n total items pick k
n! / k! ( n-k )!
