specifying the value of "n"

Question

If two coins are tossed $5$ times, then the probability of getting $5$ heads and $5$ tails is – but I got stuck in one step –

My approach: Since it seems overly similiar to binomial question so I am gonna go with that

$p$-probability of getting head on tossing a coin is $x=\text{number of times getting head}$

But what should I fill in $n$-number of trials... $5$ or $10$ and can u explain the reason?

Answer 1

I am not sure if I understand your experiment correctly:

I guess you have five turns and in each turn you throw two coins simultaneously. Your desired outcome is:

1. Turn: (head,tail)

2. Turn: (head,tail)

3. Turn: (head,tail)

4. Turn: (head,tail)

5. Turn: (head,tail)

I am not sure if you differ between the coins. Does it make a difference which coins shows which sign? In other words is

1. Turn: (tail,head)

2. Turn: (head,tail)

3. Turn: (head,tail)

4. Turn: (head,tail)

5. Turn: (head,tail)

the same result or a different? the first component refers to the first coin and respectively for the second.

I guess it is the same. So your probability to achieve (head,tail) in one turn is $p=\frac{1}{2}$ and since you have five turns you have $n=5$. By plugging that into the formula you obtain $$\binom{5}{5}\left(\frac{1}{2}\right)^5\left(\frac{1}{2}\right)^{5-5} = \left(\frac{1}{2}\right)^5$$