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

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Discussion Overview

The discussion revolves around calculating the probabilities associated with obtaining Tail/Tail outcomes when tossing two coins 20 times. Participants explore both the exact probability of achieving exactly 5 Tail/Tail outcomes and the probability of achieving at least 2 Tail/Tail outcomes. The conversation includes elements of binomial distribution and probability theory.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Homework-related

Main Points Raised

  • One participant suggests that the probability of getting Tail/Tail in a single toss is $\frac{1}{4}$, calculated as $\frac{1}{2} \cdot \frac{1}{2}$.
  • Another participant points out that the calculation for exactly 5 Tail/Tail outcomes should involve a binomial distribution formula, indicating a missing factor of 5 in the initial probability calculation.
  • A later reply proposes the binomial probability formula: $$\binom{20}{5}\cdot \left (\frac{1}{4}\right )^5\cdot \left (1-\frac{1}{4}\right )^{20-5}$$ for calculating the probability of exactly 5 Tail/Tail outcomes.
  • Participants discuss using the complement rule to find the probability of achieving at least 2 Tail/Tail outcomes, suggesting it involves calculating $1 - P(X < 2)$.
  • Specific calculations for $P(X=1)$ and $P(X=0)$ are mentioned, involving the binomial coefficients and probabilities for those outcomes.
  • One participant shares results from a simulation, reporting 202276 occurrences of Tail/Tail out of 1 million trials.

Areas of Agreement / Disagreement

Participants generally agree on the use of the binomial distribution for these calculations, but there is no explicit consensus on the final probability values, as they are still in the process of deriving them.

Contextual Notes

The discussion assumes familiarity with binomial distribution and probability calculations but does not resolve all mathematical steps or clarify all assumptions involved in the calculations.

mathmari
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Hey! :o

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)
 
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mathmari said:
Hey! :o

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.

mathmari said:
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 $5$.

mathmari said:
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)
 
I like Serena said:
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)
I like Serena said:
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)
 
Yep. All correct. (Nod)
 
mathmari said:
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.
 
Thank you! (Yes)
 

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