An irregular 6 faced dice is such that the probability that it gives 3 even numbers in 5 throws is twice the probability that it gives 2 even numbers 5 throws. How many sets of exactly 5 trials can be expected to give no even number out of 2500 sets?
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Let the probability of an even number be pp. Thus the probability of an odd number is 1−p1−p.
The probability of 33 even in 55 throws is (53)p3(1−p)2(53)p3(1−p)2. The probability of 22 even in 55throws is (52)p2(1−p)3(52)p2(1−p)3.
We are told that (53)p3(1−p)2=2(52)p2(1−p)3(53)p3(1−p)2=2(52)p2(1−p)3. Solve for pp. There are the solutions p=0p=0 and p=1p=1, which we are presumably expected to discard, although there is no mathematical reason to do so. In addition, we have the solution p=23p=23. So the probability that a single toss results in a number that is not even is 1313.
The probability of no even in 55 trials is therefore 135135.
Now in 25002500 sets of 55 trials, the expected number of times we have no even is 2500⋅1352500⋅135.
The probability of 33 even in 55 throws is (53)p3(1−p)2(53)p3(1−p)2. The probability of 22 even in 55throws is (52)p2(1−p)3(52)p2(1−p)3.
We are told that (53)p3(1−p)2=2(52)p2(1−p)3(53)p3(1−p)2=2(52)p2(1−p)3. Solve for pp. There are the solutions p=0p=0 and p=1p=1, which we are presumably expected to discard, although there is no mathematical reason to do so. In addition, we have the solution p=23p=23. So the probability that a single toss results in a number that is not even is 1313.
The probability of no even in 55 trials is therefore 135135.
Now in 25002500 sets of 55 trials, the expected number of times we have no even is 2500⋅1352500⋅135.
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