# 10 throws of coin 0-4 heads how many possibilities?

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• Feb 21st 2008, 09:48 AM
Joanna Jakma
10 throws of coin 0-4 heads how many possibilities?
If one tosses a coin ten times, how many possible outcomes are there (assuming order is important) if of the ten throws there must be between 0 -4 heads?

Not sure how to calculate that...

Eg obviously total possibilities for 10 throws in general is 2^10=1024 possibilities
• Feb 21st 2008, 09:58 AM
Plato
$\sum\limits_{k = 0}^4 {{ 10 \choose k} \left( {\frac{1}{2}} \right)^{10} }$ is the probability of no more than four heads out of ten.
• Feb 21st 2008, 11:18 AM
Joanna Jakma
REally sorry but I'm still stuck/I don't fully follow?
• Feb 21st 2008, 11:35 AM
Plato
“HHHTTTTTT” that string represents tosses of a coin where we get four heads and six tails.
That particular string has the probability of $\left( {\frac{1}{2}} \right)^{10}$ of occurring.
If we rearrange that string in anyway, say “THTHTTHHT” it still represents four heads and six tails. There are ${{10} \choose 4} = \left( {\frac{10!}{(4!)(6!)}} \right)$ way to rearrange that string.
So there are that many ways to get four heads and six tails.
In the sum we did it for 0,1,2,3, & 4.
• Feb 21st 2008, 01:01 PM
colby2152
Quote:

Originally Posted by Plato
“HHHTTTTTT” that string represents tosses of a coin where we get four heads and six tails.
That particular string has the probability of $\left( {\frac{1}{2}} \right)^{10}$ of occurring.
If we rearrange that string in anyway, say “THTHTTHHT” it still represents four heads and six tails. There are ${{10} \choose 4} = \left( {\frac{10!}{(4!)(6!)}} \right)$ way to rearrange that string.
So there are that many ways to get four heads and six tails.
In the sum we did it for 0,1,2,3, & 4.

There is probably no simpler way to explain it. Good one Plato! (Star)(Star)(Star)(Star)(Star)
• Feb 21st 2008, 01:17 PM
Soroban
Hello, Joanna!

Everyone is giving you probabilities . . .

Quote:

If one tosses a coin ten times, how many possible outcomes are there
(assuming order is important) if there must be between 0 -4 heads?

We must consider the five cases: 0 heads, 1 head, 2 heads, 3 heads, 4 heads
. . and add the results.

For 0 heads, there is one way (all Tails): $1$

For 1 head, there are: . ${10\choose1} = 10$ ways.

For 2 heads, there are: . ${10\choose2} \:=\:45$ ways.

For 3 heads, there are: . ${10\choose3} \:=\:120$ ways.

For 4 heads, there are: . ${10\choose4} = 210$ ways.

Therefore, there are: . $1 + 10 + 45 + 120 + 210 \:=\:\boxed{386\text{ ways}}$