Suppose you are playing a game in which two fair dice are to rolled. To make the first move, you need to roll doubles or a sum of 3 or 11. What is the probability that you will be able make the first move?

Answer 1

#5/18 =0.278#

so we need the probability of

#a#= rolling doubles or #b#=sum of 3 or 11.

#p(a vv b) = p(a)+p(b)# if they are independent.

In fact rolling a double will never be a sum of 3 or 11 so these are mutually exclusive events. Hence, the total numbers you can get is #6xx6# so no matter what you roll there are always #36# possible outcomes.

#p(a) = 6/36#, you can only roll a double for each number.

#p(b) = 4/36#, you can get a sum of 3 by rolling either a 1,2 or 2,1 and the same logic for obtaining an 11 from 5,6 or 6,5.

Putting it together by adding:

#p(a vv b) =6/36+4/36 = 10/36 = 5/18 =0.278#

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Answer 2

Jose Arias

The probability of rolling doubles with two fair dice is $\frac{6}{36} = \frac{1}{6}$ . The probability of rolling a sum of 3 is $\frac{2}{36} = \frac{1}{18}$ , and the probability of rolling a sum of 11 is also $\frac{2}{36} = \frac{1}{18}$ .

So, the total probability of making the first move by rolling doubles or a sum of 3 or 11 is $\frac{1}{6} + \frac{1}{18} + \frac{1}{18} = \frac{9}{18} = \frac{1}{2}$ .

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.