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.
There are 7 ways.
There are eight (8).
9, you just have to multiply for problems like this
There are 4 events: 3 heads, 2 heads 1 tail, 1 head 2 tails, and 3 tails.
n(S)=8 let A be the event that more than one tail appears n(A)=4 so,P(A)=4\8=0.5
Three half-dollars (three 50-cent coins). In US coins, a dollar and two quarters (dollar coins are not well-circulated).
There are eight (8).
Eight of them.
They are:HHH, HHT, HTH, THH, HTT, THT, TTH, TTT.
9, you just have to multiply for problems like this
2/9
It is 0.3125
TTT, TTH, THT, HTT
The chance for each toss is 1/2. The chance of all three flips being tails is 1/2 * 1/2 * 1/2 = 1/8. So the probability of at least one head is 7/8.
1/2 x 1/2 x 1/2 = 1/8 or 0.125 also 12.5%
1/2 or 50% (1/8 prob of 0H + 3/8 prob of 1H)
1/4
There are 4 events: 3 heads, 2 heads 1 tail, 1 head 2 tails, and 3 tails.