# [SOLVED]2 coins are thrown 20 times

#### mathmari

##### Well-known member
MHB Site Helper
Hey!! 2 coins are thrown 20 times. I want to calculate the probability
(a) to achieve exactly 5 times the Tail/Tail
(b) to achieve at least 2 times Tail/Tail

If we throw the 2 coins once the probability that we get Tail/Tail is equal to $\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}$, or not?
Is the probability then at (a) equal to $\left (\frac{1}{4}\right )^{20}$ ?

Could you give me a hint for (b) ? #### Klaas van Aarsen

##### MHB Seeker
Staff member
Hey!! 2 coins are thrown 20 times. I want to calculate the probability
(a) to achieve exactly 5 times the Tail/Tail
(b) to achieve at least 2 times Tail/Tail

If we throw the 2 coins once the probability that we get Tail/Tail is equal to $\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}$, or not?
Hey mathmari !!

Yep.

Is the probability then at (a) equal to $\left (\frac{1}{4}\right )^{20}$ ?
There seems to be a $5$ missing. It's a binomial distribution, so we should use the corresponding formula, which includes a spot to put the $5$.

Could you give me a hint for (b) ?
Take the complement?
It's $1$ minus the probability of getting either 0 Tail/Tail or 1 Tail/Tail. #### mathmari

##### Well-known member
MHB Site Helper
There seems to be a $5$ missing. It's a binomial distribution, so we should use the corresponding formula, which includes a spot to put the $5$.
Oh yes.. So do we have the following? $$\binom{20}{5}\cdot \left (\frac{1}{4}\right )^5\cdot \left (1-\frac{1}{4}\right )^{20-5}$$ Take the complement?
It's $1$ minus the probability of getting either 0 Tail/Tail or 1 Tail/Tail. So, we have $$P(X\geq 2)=1-P(X<2)=1-P(X=1)-P(X=0)=1-\binom{20}{1}\cdot \left (\frac{1}{4}\right )^1\cdot \left (1-\frac{1}{4}\right )^{20-1}-\binom{20}{0}\cdot \left (\frac{1}{4}\right )^0\cdot \left (1-\frac{1}{4}\right )^{20-0}$$ or not? #### Klaas van Aarsen

##### MHB Seeker
Staff member
Yep. All correct. #### Wilmer

##### In Memoriam
So do we have the following? $$\binom{20}{5}\cdot \left (\frac{1}{4}\right )^5\cdot \left (1-\frac{1}{4}\right )^{20-5}$$
Looks good.
Ran a simulation: got 202276 out of 1 million.

#### mathmari

##### Well-known member
MHB Site Helper
Thank you!! 