00:01
Okay, so for part a, similar to what we've done in previous exercises, we have that the probability of winning at least once, so win at least ones is equal to 1 minus the probability of losing n times, which is 1 minus probability of losing raise to the n, which is 1 minus the complement of the winning probability for probability of lose, raise to the n.
00:28
And so this is what we have for part a.
00:33
Now for part b, the problem.
00:35
Of ruling at least one of these cases in n rolls is 1 minus 1 minus 1 over 6 to the n which is 1 minus 5 6 to the n and the the misconception answer so the n over n answer is n over 6 so that's n over 6 over here and so the correct answers when we get when we use this expression we get that n equals 3 gives us 0 .4212.
01:17
N equals 4, or let's say n equals 6 gives us 0 .6651, and say n equals 10 is going to give us 0 .8385.
01:38
But the wrong answer, or the n over n answer, gives us for n equals 3, it gives us 3 over 6, which is 1 half.
01:47
And for n equals 6 it gives us 1 and for n equals 10 it gives us 10 over 6 which is absurd and impossible because probably cannot exceed 1 so clearly n over n is not a good approximation it's not a good approximation to 1 minus 5 over 6 or in this case we can say n over 6 now for part c we see from exercise 85 that the probability of at least one error was 0 .1052, and compared to n over 9, or n over n, i should say, and this is equal to one n's, which is approximately 0 .1111.
02:46
So, and these aren't too far off.
02:48
They're not too in commensurate.
02:52
So arguably, we can have that n over n is not necessarily a bad approximation to 1 minus 1 minus 1 over n to the n when n is very large or equivalently when p equals 1 over n is small so when this is small then this is not that bad of an approximation now for part d we have the binomial theorem can be applied here as a plus b to the n is the binomial theorem equals some from k equals 0 to n of n choose k times a to the n minus k times b to the k and this is equal to a to the n plus eight or n times a to the n minus one b plus dot dot so if we apply the binomial theorem with a equals one and b equals negative one over n then we get that one minus one over n to the n is equal to after simplification is equal to 1 minus n over n plus n choose 2 over n squared and so on, which is 1 minus n over n plus n times n minus 1 over 2 n squared plus dot dot dot.
04:30
But if n is much less than capital n, so if the little n is much less than capital n, then we have that this expression, or rather this term, this entire expression, and all of the ones after it will be negligible compared to the first two terms...