Counting Probability & Statistics Basics - 10 Coin Tosses

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In summary, there are 1024 possible sequences when a fair coin is tossed 10 times. To determine how many sequences contain exactly 3 heads, you can use combinations and the answer is 10 choose 3, or 120 possible sequences. This can be calculated by taking 10 factorial divided by 3 factorial times 7 factorial. It is recommended to have an understanding of combinations before attempting this type of problem.
  • #1
mathwurkz
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I'm taking a crack at learning probability and statistics starting from the basics. Anyways here is the question.

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A fair coin is tossed 10 times and the sequence of scores recorded.

How many sequences are there?

How many sequences are there that contain exactly 3 heads?

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The first one is no problem. I understand it. It simply is 2^10 = 1024 possible sequences.

The second one is the one i have trouble dealing with. How do you compute it with that special case added in? The answer given is 120.
 
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eg THHHTTTT... THTT...THT..TTH. your counting the exact number of sequeences that have only ("exactly") 3 HHH. not necessary in successive order(that is THHHT)
 
  • #3
mathwurkz said:
A fair coin is tossed 10 times and the sequence of scores recorded.

How many sequences are there?

How many sequences are there that contain exactly 3 heads?

For the second one, you have exactly 3 heads.

_, _, _, _, _, _, _, _, _, _

Each spot corresponds to a coin toss. First spot corresponds to the first toss, the second spot the second toss, etc.

You have 3 of the ten spots you need to put a head. One way to do this is to put a head in the first, second and third spots. Another way to do it is to place it one in the first spot one in the third spot and one in the last spot. You could list all the 120 possible ways to put 3 heads in 10 spots but there is an easier way.

If you have looked at combinations at all then by now you should see that the answer is just 10 choose 3.

[tex]
\frac{10!}{3!7!} = 120
[/tex]

If you have not looked at combinations then I would recommend you study combinations before trying to work this kind of problem out.

Best of luck,
 
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1. What is the probability of getting exactly 5 heads in 10 coin tosses?

The probability of getting exactly 5 heads in 10 coin tosses is 252/1024 or approximately 0.2461. This can be calculated using the binomial distribution formula: P(x=k) = nCk * p^k * (1-p)^(n-k) where n is the number of trials (10 in this case), k is the number of successes (5 in this case), and p is the probability of success (0.5 for a fair coin).

2. What is the probability of getting at least 8 tails in 10 coin tosses?

The probability of getting at least 8 tails in 10 coin tosses is 1 - P(x<8) = 1 - (P(x=0) + P(x=1) + P(x=2) +...+ P(x=7)). This can also be calculated using the binomial distribution formula and summing up the probabilities for each possible number of tails from 8 to 10.

3. How many different outcomes are possible in 10 coin tosses?

There are 2^10 or 1024 possible outcomes in 10 coin tosses. Each coin toss has 2 possible outcomes (heads or tails), so the total number of outcomes is the product of 2 multiplied by itself 10 times.

4. What is the expected number of heads in 10 coin tosses?

The expected number of heads in 10 coin tosses is equal to the number of trials (10) multiplied by the probability of success (0.5 for a fair coin), which is 5. This is also known as the mean or average number of heads.

5. How does the law of large numbers apply to 10 coin tosses?

The law of large numbers states that as the number of trials increases, the observed results will approach the expected results. In the case of 10 coin tosses, the more times we repeat the experiment, the closer the actual number of heads will get to the expected number of 5 heads. This can be seen by conducting multiple trials and calculating the average number of heads.

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