2.3 Independent Events

• called the Multiplication Rule for Independent Events
• makes it easy to compute P(AB)  if know just P(A), P(B)
• what does independence mean? Look at the conditional probability:

     for A and B independent,
or   P(A|B) = P(A).
• the fact that B has occurred doesn’t change the probability that A will occur
• knowledge that B has occured doesn't give us any additional information as to whether or not A might also have occurred
• A and B are independent if they "don't affect one another"

Flip a fair coin twice; use sample space  S = {HH, HT, TH, TT}  to denote result of experiment. (A fair coin is one for which the probability of getting a head or a tail on each flip is 1/2.) Compute the probability of each outcome in the sample space.

Clearly, results of flips are independent; whether or not a head occurs on the first flip has no bearing on whether or not a head will occur on the second flip.

Let

H1 = event get head on 1st flip
H2 = event get head on 2nd flip
Then
P (two heads) = P(H1H2) = P(H1) P(H2)    since H1, H2 are independent
= (1/2) (1/2)  = 1/4
 Thus
P(HH) = 1/4.
 The same reasoning follows for all other elements of the sample space; thus
P(HT) = 1/4
P(TH) = 1/4
P(TT) = 1/4
 
Thus all of the elements of the sample space are equally likely.

Consider now a weighted coin, for which the probability of getting a head is  P(H) = .7  and the probability of getting a tail is thus  P(T) = .3.  Flip this coin twice, and use sample space  S = {HH, HT, TH, TT}  to denote possible outcomes. Compute the probability of each outcome in the sample space.
As above, the results of the two flips are independent, so compute as above using the multiplication rule for independent events:
P(HH)  =  P(H) P(H)  =  (.7) (.7) =  .49,
P(HT)  =  P(H) P(T)  =  (.7) (.3)  =  .21,
P(TH)  =  P(T) P(H)  =  (.3) (.7)  =  .21,
P(TT)  =  P(T) P(T)  =  (.3) (.3)   =  .09.
 Notice that now the elements of the sample space are not equally likely!

ex:

Let R = event it rains on a given day, and W = event it’s windy. Suppose P(R) = .30, P(W) = .20, P(RW) = .15 ( = probability that it's both rainy and windy). Are these events independent?
P(R) P(W) = .06,  which is not equal to  P(R W).
Thus the events are not independent, as we might expect: if it's rainy, it's also likely to be windy.

General Multiplication Rule

If A, B independent, and know P(A) and P(B), easy to find P(A  B) from the definition of independence:

P(AB) = P(A) P(B).

P(A|B)  = 

P(AB) = P(B) P(A|B)

ex:  Urn model

Have 4 white balls, 2 black ones, in an urn.  Draw 2 balls in succession, without replacing 1st before drawing 2nd. Let the sample space be  {WW, WB, BW, BB}. Find the probability of each of the outcomes.

Note:  can't use the multiplication rule for independent events to find the results, i.e., can't use
P(WW) = P(W) P(W), since drawing a white ball on the first draw does affect probability we’ll get a white ball on the second draw (since there will thus be fewer white balls in the urn) - these events aren't independent!

Let

W1 = event 1st ball drawn is white
W2 = event 2nd ball drawn is white
B1 = event 1st ball drawn is black
B2 = event the 2nd ball drawn is black
Since these events aren't independent, use the general multiplication rule:
P(WW) = P(W1W2) = P(W1) P(W2 | W1) = (4/6) (3/5) = 12/30
in this computation, P(W1) = 4/6, since initially there are 4 white balls out of 6 total in the urn; but then P(W2|W1) = 3/5: the probability of getting a white ball on the second draw, given that we got a white ball on the first draw, is 3/5, since there are only 5 balls left in the urn, of which 3 are white.
Similarly,
P(WB) = P(W1B2) = P(W1) P(B2 | W1) = (4/6) (2/5) = 8/30
P(BW) = P(B1W2) = P(B1) P(W2 | B1) = (2/6) (4/5) = 8/30
P(BB) = P(B1B2) = P(B1) P(B2 | B1) = (2/6) (1/5) = 2/30

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