MHB How to Calculate Tail/Tail Probability for Multiple Coin Tosses?

[mathmari]

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) ? (Wondering)

 

[I like Serena]

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. (Worried)
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. (Thinking)

 

[mathmari]

There seems to be a $5$ missing. (Worried)
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? $$\binom{20}{5}\cdot \left (\frac{1}{4}\right )^5\cdot \left (1-\frac{1}{4}\right )^{20-5}$$ (Wondering)

Take the complement?
It's $1$ minus the probability of getting either 0 Tail/Tail or 1 Tail/Tail. (Thinking)

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? (Wondering)

 

[I like Serena]

Yep. All correct. (Nod)

 

[Wilmer]

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]

Thank you! (Yes)