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Ignoring the very small possibility that a coin tossed will land on its edge and stay there, I'll assume each coin can land either Head or Tails.

The reverse of the condition "at least one Head" is "no Heads"

Thus the number of ways of at least one Head is the same as the total number of ways less the number of way of no Heads.

There are 2 × 2 × 2 = 2³ = 8 possible ways the three coins can land.

Of these there is only 1 way which shows no Heads: TTT

Thus there are 8 - 1 = 7 ways that at least one Head shows

You could list all the possible ways that at least one head shows (which are: HTT, THT, TTH, HHT, HTH, THH, HHH - 7 of them), but the "subtractive" method above is much quicker and easier and less prone to error in the listing.

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7y ago

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