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The easiest way to compute the probability in question is through the “1 − x” shortcut. To do so, imagine the opposite of the event of interest, namely, that none of the n throws yields a 6. The probability of a single throw not yielding a 6 is 5/6, and because each throw is independent, the cumulative probability of none of the n throws yielding a 6 is found by multiplication:

P(n 6 in n throws)=(5/6)^n

Powers of fractions less than one get smaller as the exponent increases. Thus, this probability will become very small for large values of n, such that the probability of getting at least one 6 (which is 1-(5/6)^n ) will come closer and closer to 1. Thus, as n increases, it becomes more and more certain that a 6 will be thrown. The question now is, What is the smallest that the probability of getting at least one six could be? To answer that question, you should set n to its lowest possible value, which is 3. In that case, the probability of never getting a 6 is given by:

(5/6)^3=125/216

However, the probability of getting at least one 6 in three throws is given by:

P (At least 1/6 in three throws ) = 1-125/256=91/216

This value is less than 1/2. As you saw earlier, however, as n grows, it becomes ever more likely that at least one throw will yield a 6, so that the probability eventually surpasses 1/2. Thus, Quantity A can be less than or greater than 1/2. Thus, the relationship cannot be determined from the information given.