2 coins are thrown 20 times

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  • Thread starter mathmari
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In summary, the conversation discusses calculating the probability of getting a certain number of Tail/Tail outcomes when throwing 2 coins 20 times. The probability is determined using the binomial distribution formula, and in the end, a simulation is also used.
  • #1
mathmari
Gold Member
MHB
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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)
 
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  • #2
mathmari said:
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.

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)
 
  • #3
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)
 
  • #4
Yep. All correct. (Nod)
 
  • #5
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.
 
  • #6
Thank you! (Yes)
 

1. What is the probability of getting two heads in a row?

The probability of getting two heads in a row when throwing 2 coins 20 times is 1/4 or 25%. This is because the probability of getting a head on the first throw is 1/2 and the probability of getting a head on the second throw is also 1/2. Therefore, the probability of getting two heads in a row is 1/2 x 1/2 = 1/4 or 25%.

2. What is the expected number of heads when throwing 2 coins 20 times?

The expected number of heads when throwing 2 coins 20 times is 10. This is because the probability of getting a head on each throw is 1/2, so out of 20 throws, we can expect to get 10 heads.

3. What is the probability of getting at least one tail when throwing 2 coins 20 times?

The probability of getting at least one tail when throwing 2 coins 20 times is 99.99999999999997%. This means that there is a very high chance of getting at least one tail when throwing 2 coins 20 times. In fact, the probability of getting all heads is extremely low (0.00000000000003%), so it is almost certain that at least one tail will appear.

4. How many possible outcomes are there when throwing 2 coins 20 times?

There are 2^20 or 1,048,576 possible outcomes when throwing 2 coins 20 times. This is because each throw has 2 possible outcomes (heads or tails) and there are 20 throws. So, the total number of possible outcomes is 2 x 2 x 2 x ... (20 times) = 2^20 or 1,048,576.

5. What is the significance of throwing 2 coins 20 times in terms of probability and statistics?

Throwing 2 coins 20 times is a simple and common example used to demonstrate the principles of probability and statistics. It allows us to calculate and analyze the probability of different outcomes and understand concepts such as expected value, independence, and randomness. Additionally, it can be used to illustrate the law of large numbers, which states that as the number of trials increases, the observed outcomes tend to approach the expected outcomes.

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