5 coins are thrown. It came up H 3 times, and T twice. what is the probability that the first throw was H? I've got the following question:

5 coins are thrown. It came up H 3 times, and T twice. what is the probability that the first throw was H?

What is the formula I need to use to solve it?
 A: The total number of ways to get $3$ heads and $2$ tails is $\binom53=\binom52=10$
The number of ways to get head first, then $2$ heads and $2$ tails is $\binom42=6$
Hence the probability is $\frac{6}{10}$

You can also calculate it by counting down the options (first $6$ out of $10$):


*

*HHHTT

*HHTHT

*HHTTH

*HTHHT

*HTHTH

*HTTHH

*THHHT

*THHTH

*THTHT

*TTHHH

A: I am assuming the coin is fair.
There are a simpler way and a harder way of solving this problem. The simpler way is to observe that 3 out of 5 throws were heads and therefore if we order the throws randomly, we are effectively choosing a random throw to be first so the probability of it being heads is $\frac{3}{5}$.
The harder way is as follows. There are ${5 \choose 3} = 10$ orderings of the heads and tails and ${4 \choose 2} = 6$ orderings where the first throw was heads. Therefore the chance to get heads on the first throw is $\frac{6}{10} = \frac{3}{5}$.
A: The events are not linked. The first throw being heads does not make it more likely for the second one to be H or T.
Out of the 5 throws, 3 were heads.
Because they are not linked events, the first being H is the same as another being H, or $\frac{3}{5}$
A: $3$ letters $H$ and $2$ letters $T$ are to be placed on $5$ spots. What is the probability that spot number $1$ will receive an $H$? $$\frac{3}{5}$$
