exactly how would I do this question? I understand if the concern said: at the very least one head then I would do:

\$5choose0=1\$

\$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:

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\$.

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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?

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reply Jul 11 "13 in ~ 0:15 Zev ChonolesZev Chonoles
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