# Classical Probability Model

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• Sep 29th 2009, 04:45 PM
gconfused
Classical Probability Model
Hi, well I'm totally lost with this subject, and this is an easy question but I just don't understand ..
Question: A fair coint is flipped n = 100 times. What is the probability to have exactly k = 50 heads? What is the probability to have exactly n/2 heads when n is even?

Answer:
Use the Classical Probability Model: $P(E) = \frac{N(E)}{N(S)}$ with $N(S) = 2^{100}$

<I don't understand why it is $2^{100}$. Why isn't it just 100? >

So $P(E) = \frac{100!}{50!50!} \times 2^{-100}$

<Again, I don't get why we're using the combination formula? Why isn't it just $\frac{50}{100}$...????>

Can someone please explain this to me?
• Sep 29th 2009, 10:30 PM
noname
For the first question you should use the binomial distribution with n = 100, k = 50 and p = 1/2

n is the number of bernoulli trial
k is the number of success
p is the probability of success on a single trial, it's always the same because the outcome of every trial is independent from the others

Hint: if n is even you can write it m*2
• Sep 30th 2009, 02:06 PM
awkward
Quote:

Originally Posted by gconfused
Hi, well I'm totally lost with this subject, and this is an easy question but I just don't understand ..
Question: A fair coint is flipped n = 100 times. What is the probability to have exactly k = 50 heads? What is the probability to have exactly n/2 heads when n is even?

Answer:
Use the Classical Probability Model: $P(E) = \frac{N(E)}{N(S)}$ with $N(S) = 2^{100}$

<I don't understand why it is $2^{100}$. Why isn't it just 100? >

So $P(E) = \frac{100!}{50!50!} \times 2^{-100}$

<Again, I don't get why we're using the combination formula? Why isn't it just $\frac{50}{100}$...????>

Can someone please explain this to me?

If you flip a coin 100 times and write down the outcome of each flip, there are $2^{100}$ possible sequences, each of which is equally likely. How many of those have exactly 50 heads? Exactly as many as there are ways to pick 50 flips out of the sequence of 100, i.e., the number of combinations of 100 objects taken 50 at a time, $\binom{100}{50} = \frac{100!}{50! \; 50!}$.