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

Question

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?

Answer 1

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 $\frac{6}{10} = \frac{3}{5}$.

Answer 2

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 $\frac{6}{10}$

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

1. HHHTT
2. HHTHT
3. HHTTH
4. HTHHT
5. HTHTH
6. HTTHH
7. THHHT
8. THHTH
9. THTHT
10. TTHHH

Answer 3

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 $\frac{3}{5}$

Answer 4

$3$ letters $H$ and $2$ letters $T$ are to be placed on $5$ spots. What is the probability that spot number $1$ will receive an $H$? $$\frac{3}{5}$$