Counterintuitive Probability Problem

[SprucerMoose]

Homework Statement

A fair coin is tossed 4 times.
The following events are defined:
A = Exactly 2 tosses are heads
B = The second toss is a head

Are events A and B independent?

The Attempt at a Solution

My workbook says the events are independent, but it's just seems so counterintuitive.

Events for A
HHTT HTHT HTTH
TTHH THHT THTH

Pr(A) = 6/16 = 3/8

Pr(B) = 1/2

Events for A ∩ B
HHTT THHT THTH

Pr(A ∩ B) = 3/16

Pr(A) x Pr(b) = (3/8) x (1/2) = 3/16

Thus Events A and B are independent by my textbooks definition.

Mathematically these events appear to demonstrate that they are independent as Pr(A) x Pr(b) = Pr(A ∩ B), but I just can't see how the occurrence of event B does not affect event A. Knowing that B occurs does not change the probability of A occurring as there are half as many favourable outcomes and also half as many total outcomes. So does this mean that even though event B's occurence affects the favourabe outcomes of A, it is still technically independent of A because the numerical probability is unchanged?

Any comments would be appreciated.

 

[HallsofIvy]

But your computation, which agrees with mine, shows that B's occurence does NOT affect the favorable outcomes of A.

There are 16 possible outcomes when flipping four coins and 6 of them are "favorable" for A. If we fix the second coin as a head, there are not 8 possible outcomes and 3 of them of them are "favorable". While the specific outcomes have changed, the ratio has not: 6/16= 3/8 and that is all that matters. Yes, it is the "numerical probability" that counts. And that's a good thing- otherwise we could not use the many simplifying formulas to calculate probabilities!

 

[SprucerMoose]

Thanks very much for the help. That question has been bothering me all day. Your assistance is greatly appreciated.

 

[tiny-tim]

Hi SprucerMoose!

Here's a slightly different way of looking at it …

Mathematically these events appear to demonstrate that they are independent as Pr(A) x Pr(b) = Pr(A ∩ B) …

I agree that it's difficult to see why A and B, in ordinary English, should be called independent with that definition.

But it makes more sense if you use the equivalent definition:

P(A|B) = P(A ∩ B)/P(B)

P(B|A) = P(A ∩ B)/P(A)

(btw we usually just use "P" ) …

the probability of getting the result of one is not affected by knowing the result of the other

 

[SprucerMoose]

Thanks again