because night and time together will equal nighttime.
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t(1) = 1 t(2) = 1 t(n+1) = t(n) + t(n-1) for n = 1, 2, 3, ... That is, the first two terms are 1; after that every term is the sum of the previous two terms.
There are 8 permutations of three coins. Of these, 3 of them have two heads, so the probability of tossing two heads on three coins is 3 in 8, or 0.375. However, you said, "at least", so that includes the case of three heads, so the probability of throwing at least two heads is 4 in 8, or 0.5. T T T T T H T H T T H H * H T T H T H * H H T * H H H *
The intersection of two sets S and T is the set of all elements that belong to both S and T.
4 : Each coin toss multiplies the generated outcome by two. The outcome for two tosses can be Heads (H) Tails (T), H H, T T or T H. As such, with three tosses the outcome can be H H H, H H T, H T H, T H H, T T H, T T T, T H T, H T T. We can define the number of outcomes as 2^n, where n is the amount of tosses.