if three coins are tossed at the same time, what is the probability that at least one of the coins will land heads up. The answer is 7/8 but explain why:confused:

all the possible outcomes are:

HHH
HHT
HTH
THH
HTT
THT
TTH
TTT

where H is heads and T is tails

of the 8 possible outcomes only one has no heads up toss and the remaining 7 do

so the probability that at least one of the coins will land heads up. The answer is 7/8

I'm not sure what level you are working at, so let me know if that doesn't make sense to you...

Of the 8 possible outcomes only one has no heads up toss and the remaining 7 do, so the probability that at least one of the coins will land heads up is 7/8

Or, if you want to use that method:

Probability of having all tails (in all the other outcomes, there is at least one head)

= \frac{1}{2} \times\frac{1}{2} \times\frac{1}{2}

= \frac{1}{8}

Then, to have your answer, do:

= 1 - \frac{1}{8}

= \frac{7}{8}

Hope it helped!