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[SOLVED] 2 coins are thrown 20 times

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,004
Hey!! :eek:

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)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,684
Hey!! :eek:

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

Well-known member
MHB Site Helper
Apr 14, 2013
4,004
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)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,684
Yep. All correct. (Nod)
 

Wilmer

In Memoriam
Mar 19, 2012
376
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
Apr 14, 2013
4,004
Thank you!! (Yes)