How Do You Calculate Probabilities for Multiple Coin Tosses?

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To calculate probabilities for multiple coin tosses, start by listing all possible outcomes, which total eight for three coins. For the event of all coins showing heads, there is only one favorable outcome. To find the probability of at least one tail, consider the complementary event of all heads, which simplifies the calculation. For exactly two heads, count the favorable outcomes directly from the list. Understanding the principles of independent events and using addition and subtraction for overlapping probabilities is crucial for accurate calculations.
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Im just starting probability and need some help. I have a question in my text that i can't figure out.

Three coins are tossed at the same time. Find the probability that
a) all come up heads
b) at least one comes up tails
c) exactly two come up heads


I know that the total possible outcomes is 8 but i don't know how to figure out the outcomes of the event.


Thanks for any help!

Adam
 
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Hi, since there are only 8 possible outcomes, why don't you start by writing down all of them and see how often each event occurs? Once you've done that, see if you can come up with a more clever approach that will work for more general circumstances.
 
For the first part,i'm sure you realize that there are independent events involved...
What is the number of favorable outcomes for points b) and c)...?

Daniel.

P.S.The problem is so simple,you can even count them...But that's only for checking your logics...It could have been 100 coins being flipped at the same time... :-p
 
Yeah I figured it out after i posted the question...oh well.

Thanks anyways
 
Remember, if multiple events need to cooexist, multiply their probabilities together (similar to an "AND" gate in programming) and if either goes, add their probabilites (similar to "OR" gates)
 
The Divine Zephyr said:
Remember, if multiple events need to cooexist, multiply their probabilities together (similar to an "AND" gate in programming) and if either goes, add their probabilites (similar to "OR" gates)

No no no. If either goes, add their probabilties and subtract the AND. Otherwise you count the center twice.

A and B = A + B - (A or B)
Check out a Venn diagram for a better idea of what's going on.
 
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