You are watching: If two coins are tossed in the air, what is the probability that both coins will land on "heads"?

But someone rather tells me that together the possible events are:

Head - HeadHead - TailTail - HeadTail - Tailthen once we gain a head we restrict ourselves come the an initial three cases, therefore the probability would be $frac13$.

What is the appropriate way?

I know there"s a difference between saying "first come head" come say "one of the two came head", but if we have actually the first fact, aren"t we supposed to recognize which one is that come head?

probability combinatorics discrete-mmsanotherstage2019.comematics

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edited Feb 13 "17 at 14:21

MR_BD

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asked may 30 "16 at 4:37

Ronald BecerraRonald Becerra

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$egingroup$ Um, "at least one gets head" can additionally be "interpreted multiple ways", if you recognize what ns mean... Normally one states of a coin the it "comes up heads". $endgroup$

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may 31 "16 in ~ 3:09

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The problem here is that your expression "one it s okay head" is no

**precise**enough.

If it method "at the very least one of the coins comes up heads", then certainly there room 3 equally most likely possibilities (HT,TH,HH) out of which precisely 1 has actually both comes up heads. This way that the wanted probability is $frac13$.

If it method "out of the two coins A,B the were flipped, A comes up heads", climate B is equally most likely to come up heads or tails, and also so the wanted probability is $frac12$.

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answered may 30 "16 in ~ 7:25

user21820user21820

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I think this is one of the instances where logic/mmsanotherstage2019.comematics entirely goes overboard top top a trivial problem, and you (as well as your friend) are overthinking it. But it"s fun anyway, so...

There space at least 3 possible answers which room equally correct, depending on how pedantically one tries come twist the wording one way or the other. But for handy purposes the does no make a many sense due to the fact that there is only exactly one solution that is correct (and instantly obvious) in every situation.

The an initial obvious translate is the you toss 2 coins right into the air in ~ the very same time (which is *not* what girlfriend describe!). There room 4 possible outcomes. Among these outcomes has actually two heads, and also one has no heads at all. Girlfriend have set up the precondition that one coin gets head, which rules the end the "no heads" outcome, leaving 3 feasible outcomes. Only among the three has two heads in it, thus: 1/3. This is a *dependent* probability. That is also an instance of a Monty hall Problem.

The second obvious interpretation is friend toss one coin, and it come up head. That"s the precondition. You might just as well not have actually tossed the an initial coin in ~ all. You now toss the second coin. Alternatively, you deserve to toss the two coins together, but ignore all instances where the precondition that the very first one gets head isn"t fulfilled.Assuming the second coin is not weighted or a cheat coin v two top or such, the opportunity is, that course, 1/2. Native the point of watch of the 2nd coin, the an initial coin doesn"t exist at all. This is a solitary (independent) probability.

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The third obvious solution is zero. If 2 coins space flipped and *exactly* one coin it s okay head, the probability the both coins acquiring head is zero. This is a *smart ass* probability.