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Experimental probability depends on each trial. It is not known until you conduct the trial. You are probably talking about theoretical probability in the question.

The solve this problem, simplify it...

The probability of rolling a 1 or a 2 in a standard six sided die is 1 in 3, or about 0.3333. Remember this, because I'm going to change the problem.

Think about throwing a coin. What is the probability of throwing 39 heads in 100 coin tosses? Its the same problem, and I'll show this at the end.

The probability of throwing a head is 1 in 2, or 0.5. The probability of throwing 39 heads in 100 coin tosses is more difficult.

The real question is, what is the number of combinations of 100 things taken 39 at a time? Sound familiar? It better. Each of those 39 events corresponds to a head, and the 100 things is the number of coin tosses. The answer is that the number of combinations of 100 things taken 39 at a time is 100! / (39! (100-39)!), or about 9.0139 x 1027.

Since probability is the number of permutations of the desired outcome divided by the number of permutations of the possible outcome, simply divide the above number by 100!, or 9.3326 x 10157, getting about 9.6585 x 10-131.

Now, back to the original problem. We were talking about a six-sided die, and wanting a sample space of 6, but coins have a a sample space of 2. Reconsider and you find that the actual sample space is 3, because we want a 1 or a 2, so multiply 9.6585 x 10-131 by 2/3 to get 6.439 x 10-131. That is the probability of rolling a 1 or a 2 exactly 39 times in 100 rolls.

From Rafaelrz:

You have two possible events in a single throw of the die: A, die comes out 1 or 2; B, die comes out 3,4,5,or 6.

Probability of events are; P(A) is 1/3, P(B) is 2/3.

If you want 39 'A' events in a trial of 100 throws, the number of ways this events

can happen ( in which number of throw they appear ) is given by the number of

combinations above calculated nCr (n is 100, r is 39) with the result of about

9.01392403 x 1027.

Then, the probability of getting 39 'A' events in a trial of 100 throws is:

100C39 x [P(A)]39 x [P(B)]61 equal to 9.01392403 x 1027 x (1/3)39 x (2/3)61

equal to 0.040329065 ( Theoretical probability )

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Q: What is the experimental probability of rolling one die 100 times and getting a 1 or a 2 39 times?
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