Union of probabilities(coin tosses)

  • Thread starter Thread starter torquerotates
  • Start date Start date
  • Tags Tags
    Union
torquerotates
Messages
207
Reaction score
0

Homework Statement


Toss a coin 3 times. What is the probability that we get a head on the first toss or a head on the second toss or a head on the third toss?


Homework Equations


Pr(AorB)=Pr(A)+Pr(B)-Pr(AandB)



The Attempt at a Solution


A=head on 1st toss
B=head on 2nd toss
C=head on 3rd toss

Pr(AorBorC)=Pr(A)+Pr(B)+Pr(C)-Pr(A&B&C)

=(1/2)+(1/2)+(1/2)-(1/8)=11/8

but 11/8>1 which is a contradiction. So something is wrong with my sol.
 
Physics news on Phys.org
torquerotates said:

Homework Statement


Toss a coin 3 times. What is the probability that we get a head on the first toss or a head on the second toss or a head on the third toss?


Homework Equations


Pr(AorB)=Pr(A)+Pr(B)-Pr(AandB)

This formula is true



The Attempt at a Solution


A=head on 1st toss
B=head on 2nd toss
C=head on 3rd toss

Pr(AorBorC)=Pr(A)+Pr(B)+Pr(C)-Pr(A&B&C)

This one isn't. The formula above only works as written when you have precisely two sets. You can expand it to a more general setting by considering what it means though. P(A or B) = P(A) + P(B) - P(A and B)

What this is really doing is just a counting argument. The number of things in the two sets A and B is just the number of things in A plus the number of things in B minus the number of things in both A and B. We had to subtract the last part because everything that was in A and B was counted when we counted the number of things in A, and also counted when we counted the number of things in B.

So similarly, to find P(A or B or C), we need the number of things in A, plus the number of things in B, plus the number of things in C. But in this case we double counted everything that is in A and B, and we double counted everything in B and C, and we double counted everything in A and C for the same reason as above. Also, anything in all three of A and B and C was counted three times, but when we remove everything in A and B, and everything in A and C, and everything in B and C, we uncounted those things three times also. So we need to add back in the number of things in all of A, B and C at the same time.

Do you see how to transform that into a formula about probabilities?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top