$2^5=32-1 = 31 $
You"re correct that there are $;2^5 = 32$ possible outcomes of tossing 5 coin.
There space $inom51 = 5$ of these outcomes which contain specifically one head. Indeed, these possible outcomes are specifically those detailed below:
$(1)quad H;T;T;T;T$$(2)quad T;H;T;T;T$$(3)quad T;T;H;T;T$$(4)quad T;T;T;H;T$$(5)quad T;T;T;T;H$
That gives us a probability of $;dfrac532;$ that exactly one head will face up upon tossing $5$ fair coins.
You are watching: What is the probability of getting zero heads in five tosses?
We assume the the coin is fair and also is flipped fairly.
There are $2^5$ equally likely strings of size $5$ consisted of of the letter H and/or T.
There are precisely $5$ strings the have specifically $1$ H and $4$ T.
So the required probability is $dfrac52^5$.
Remark: intend that a coin has probability $p$ of landing heads, and also $1-p$ that landing tails. If the coin is tossed separately $n$ times, then the probability of specifically $k$ top is $inomnkp^k(1-p)^n-k$.
In our case, $n=5$, $p=1/2$, and also $k=1$.
edited Jul 11 "13 in ~ 0:21
answered Jul 11 "13 in ~ 0:14
André NicolasAndré Nicolas
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Hint: How many ways room there of selecting which coin out of the five will come up "head"? just how many possible outcomes space there total?
reply Jul 11 "13 in ~ 0:15
Zev ChonolesZev Chonoles
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