Yes.
Given two events, A and B, Pr(A and B) = Pr(A)*Pr(B) if A and B are independent and Pr(A and B) = Pr(A | B)*Pr(B) if they are not.
y=mx+b is the equation for a linear relationship. y= the dependant variable m= the slope of the line x= the independent variable b= the y-intercept
the circles do not overlap at all.
P(A given B)*P(B)=P(A and B), where event A is dependent on event B. Finding the probability of an independent event really depends on the situation (dart throwing, coin flipping, even Schrodinger's cat...).
Yes.
Given two events, A and B, Pr(A and B) = Pr(A)*Pr(B) if A and B are independent and Pr(A and B) = Pr(A | B)*Pr(B) if they are not.
p(A and B) = p(A) x p(B) for 2 independent events p(A and B and ...N) = p(A) x p(B) x p(C) x ...x p(N) In words, if these are all independent events, find the individual probabilities if each and multiply them all together.
if P(A)>0 then P(B'|A)=1-P(B|A) so P(A intersect B')=P(A)P(B'|A)=P(A)[1-P(B|A)] =P(A)[1-P(B)] =P(A)P(B') the definition of independent events is if P(A intersect B')=P(A)P(B') that is the proof
.7
when the occurance of an event B is not affected by the occurance of event A than we can say that these events are not dependent with each other
y=mx+b is the equation for a linear relationship. y= the dependant variable m= the slope of the line x= the independent variable b= the y-intercept
the circles do not overlap at all.
first prove *: if A intersect B is independent, then A intersect B' is independent. (this is on wiki answers) P(A' intersect B') = P(B')P(A'|B') by definition = P(B')[1-P(A|B')] since 1 = P(A) + P(A') = P(B')[1 - P(A)] from the first proof * = P(B')P(A') since 1 = P(A) + P(A') conclude with P(A' intersect B') = P(B')P(A') and is therefore independent by definition. ***note*** i am a student in my first semester of probability so this may be incorrect, but i used the first proof* so i figured i would proof this one to kinda "give back".
Answer: The property that is illustrated is: Symmetric property. Step-by-step explanation: Reflexive property-- The reflexive property states that: a implies b Symmetric Property-- it states that: if a implies b . then b implies a Transitive property-- if a implies b and b implies c then c implies a Distributive Property-- It states that: a(b+c)=ab+ac If HAX implies RIG then RIG implies HAX is a symmetric property.
P(A given B)*P(B)=P(A and B), where event A is dependent on event B. Finding the probability of an independent event really depends on the situation (dart throwing, coin flipping, even Schrodinger's cat...).
If two events are disjoint, they cannot occur at the same time. For example, if you flip a coin, you cannot get heads AND tails. Since A and B are disjoint, P(A and B) = 0 If A and B were independent, then P(A and B) = 0.4*0.5=0.2. For example, the chances you throw a dice and it lands on 1 AND the chances you flip a coin and it land on heads. These events are independent...the outcome of one event does not affect the outcome of the other.