Two
2x2x2=8 possible outcomes. In general for n tosses there are 2^n outcomes.
there can be only three combo's ------ head n tail,,,,,,,, tail n tail,,,,,,,,,,,,,, head n head
That depends what coins are allowed. Since I don't know in what country you live, there is really no way to know. For "n" different coins, you have "n" options where the same coin is used twice; you also have n(n-1)/2 options that use two different coins.
The answer is not easy: it depends on the situation that you are studying. It is simple to identify all the possible outcomes of rolling a die and tossing a coin: it is the set of ordered pairs defined as the set {(n, a), where n is an integer from 1 to 6 and a is H or T}. In general, it is not at all simple to determine all possible outcomes. There may be outcomes that you did not expect or and make allowance for. In practise, therefore, it is sometimes best to have a category for "other outcomes". In principal, a tossed coin could end up on its edge, or at least, end up leaning against some obstruction so that it is not flat on H or T.
M=0 n=0 m*n=0
(0.5)n
There are three possibilities, Heads, Tails and stand on edge
2x2x2=8 possible outcomes. In general for n tosses there are 2^n outcomes.
In three tosses there can be four possible outcomes: three heads, three tails, two heads and one tail, and one head and two tails. ^^^That is wrongA coin is tossed N times. There are 2 possibilities when you toss a coin: heads and tails.So the formula is 2^N (thats to the N power)A coin tossed 3 times has 2^3=8 possible outcomes:head head headhead head tailhead tail headtail head headtail tail headhead tail tailtail head tailtail tail tailThere they are!
n(S)=8 let A be the event that more than one tail appears n(A)=4 so,P(A)=4\8=0.5
There are 2^n elements, where n is the number of coins.
there can be only three combo's ------ head n tail,,,,,,,, tail n tail,,,,,,,,,,,,,, head n head
That depends what coins are allowed. Since I don't know in what country you live, there is really no way to know. For "n" different coins, you have "n" options where the same coin is used twice; you also have n(n-1)/2 options that use two different coins.
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The answer is not easy: it depends on the situation that you are studying. It is simple to identify all the possible outcomes of rolling a die and tossing a coin: it is the set of ordered pairs defined as the set {(n, a), where n is an integer from 1 to 6 and a is H or T}. In general, it is not at all simple to determine all possible outcomes. There may be outcomes that you did not expect or and make allowance for. In practise, therefore, it is sometimes best to have a category for "other outcomes". In principal, a tossed coin could end up on its edge, or at least, end up leaning against some obstruction so that it is not flat on H or T.
M=0 n=0 m*n=0