Union of probabilities(coin tosses)

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SUMMARY

The probability of getting at least one head when tossing a coin three times can be calculated using the principle of inclusion-exclusion. The correct formula for three events A, B, and C (representing heads on the first, second, and third tosses respectively) is Pr(A or B or C) = Pr(A) + Pr(B) + Pr(C) - Pr(A and B) - Pr(B and C) - Pr(A and C) + Pr(A and B and C). The initial attempt incorrectly calculated the probability as 11/8, which is impossible, highlighting the need for proper application of the inclusion-exclusion principle.

PREREQUISITES
  • Understanding of basic probability concepts
  • Familiarity with the inclusion-exclusion principle
  • Knowledge of event notation in probability (e.g., A, B, C)
  • Ability to perform calculations with fractions
NEXT STEPS
  • Study the inclusion-exclusion principle in depth
  • Learn how to calculate probabilities for multiple events
  • Explore combinatorial counting methods in probability
  • Practice problems involving multiple coin tosses and their outcomes
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Students studying probability theory, educators teaching probability concepts, and anyone interested in understanding the complexities of calculating probabilities for multiple events.

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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.
 
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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?
 

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