C784 Statistics Formulas
Two Events are Independent If:
P(A and B) = P(A) · P(B)
P(A | B) = P(A)
P(B | A) = P(B)
P(B | A) = P(B | not A) Two Events are Independent If:
P(A and B) = P(A) · P(B)
P(A | B) = P(A)
(it does not matter if B has occurred or not)There are two outcomes, TH and HH, that comprise event B
(tossing a H on the second throw). One of those outcomes, HH, is included in event A (tossing a H on the
first throw). Therefore the probability that A occurs given B has occurred is 1/2 (one out of 2 chances)
P(A)= 1/2 There are two outcomes that comprise event A: HT and HH. There are four total outcomes.
P(B | A) = P(B)
(for the probability of B, it does not matter whether A has occurred) The logic here is the same as that
used to calculate P(A | B). There are two outcomes, HT and HH, that are included in event A (tossing a H
on the first throw). One of those outcomes, HH, satisfied event B (tossing a H on the second throw).
Therefore the probability that B occurs given A has occurred is 1/2
P(B)= 1/2. There are two outcomes with which B can occur, TH and HH; there are four outcomes total
, P(B | A) = P(B | not A)
P(not A) has two possible outcomes, TH and TT. Of those two outcomes, only one satisfies event B
occurring, TH. Therefore the probability is 1/2
P(B | A) = 1/2
P(B | not A) = 1/2
Probability of independent Events P(A and B)
P(A and B) = P(A) x P(B) independent Events
P(A and B)
P(A and B) = P(A) x P(B)
Probability of independent Events P(A or B)
P(A or B) = P(A) + P(B) - P(A and B) independent Events P(A or B)
P(A or B) = P(A) + P(B) - P(A and B)
DISJOINT EVENTS
Probability of independent Events P(A or B)
P(A or B) = P(A) + P(B) DISJOINT EVENTS
independent Events P(A or B)
Two Events are Independent If:
P(A and B) = P(A) · P(B)
P(A | B) = P(A)
P(B | A) = P(B)
P(B | A) = P(B | not A) Two Events are Independent If:
P(A and B) = P(A) · P(B)
P(A | B) = P(A)
(it does not matter if B has occurred or not)There are two outcomes, TH and HH, that comprise event B
(tossing a H on the second throw). One of those outcomes, HH, is included in event A (tossing a H on the
first throw). Therefore the probability that A occurs given B has occurred is 1/2 (one out of 2 chances)
P(A)= 1/2 There are two outcomes that comprise event A: HT and HH. There are four total outcomes.
P(B | A) = P(B)
(for the probability of B, it does not matter whether A has occurred) The logic here is the same as that
used to calculate P(A | B). There are two outcomes, HT and HH, that are included in event A (tossing a H
on the first throw). One of those outcomes, HH, satisfied event B (tossing a H on the second throw).
Therefore the probability that B occurs given A has occurred is 1/2
P(B)= 1/2. There are two outcomes with which B can occur, TH and HH; there are four outcomes total
, P(B | A) = P(B | not A)
P(not A) has two possible outcomes, TH and TT. Of those two outcomes, only one satisfies event B
occurring, TH. Therefore the probability is 1/2
P(B | A) = 1/2
P(B | not A) = 1/2
Probability of independent Events P(A and B)
P(A and B) = P(A) x P(B) independent Events
P(A and B)
P(A and B) = P(A) x P(B)
Probability of independent Events P(A or B)
P(A or B) = P(A) + P(B) - P(A and B) independent Events P(A or B)
P(A or B) = P(A) + P(B) - P(A and B)
DISJOINT EVENTS
Probability of independent Events P(A or B)
P(A or B) = P(A) + P(B) DISJOINT EVENTS
independent Events P(A or B)
