Thread: 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

is the probability of no more than four heads out of ten.

REally sorry but I'm still stuck/I don't fully follow?

“HHHTTTTTT” that string represents tosses of a coin where we get four heads and six tails.
That particular string has the probability of of occurring.
If we rearrange that string in anyway, say “THTHTTHHT” it still represents four heads and six tails. There are 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.

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 of occurring.
If we rearrange that string in anyway, say “THTHTTHHT” it still represents four heads and six tails. There are 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!

Hello, Joanna!

Everyone is giving you probabilities . . .

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):

For 1 head, there are: . ways.

For 2 heads, there are: . ways.

For 3 heads, there are: . ways.

For 4 heads, there are: . ways.


Therefore, there are: .