When drawing this problem out visually, I got 8/12 not 7/12 as it describes in the back of the book. Did I over count for numbers that can have a coin tossed twice? Help would be much appreciated :)
weisbart: Feb. 8, 2015, 11:15 a.m.
Probably. It seems that you have the denominator right. Remember that in the first case, the probability that you toss a head (at least one) is $\frac{3}{4}$. You have four possible outcomes and three have at least one head. In the second case (roll 3,4,5,6) you have a $\frac{1}{2}$ chance of getting a head. You have to weight both cases equally, so you consider $\frac{1}{2}$ to be $\frac{2}{4}$. Now both cases have four outcomes, in the first, three lead to success and in the second, two lead to success. Your (fake) sample set will have $4\times 6$, or 24 outcomes. $2\times 3 + 4\times 2$, or 14 will lead to having tossed at least one head. This gives a probability of $\frac{14}{24} = \frac{7}{12}$.
AshleyChang: Feb. 8, 2015, 12:04 p.m.
I'm a bit confused as to how we got the fake sample set. I got the correct answer, but with a different way of thinking...Where did the 4 come from in 4 x 6? And why 2x3 + 4x2?
weisbart: Feb. 8, 2015, 3:11 p.m.
I should first remind everyone reading that I don't think that fake sample spaces are a nice way to solve problems. We'll learn a much better way on Wednesday. \[\] The set of outcomes has the following structure. It is partitioned into two subsets. For the first subset, we have a choice of 1 or 2 followed by a second choice of hh, ht, th, or tt. For the second subset, we have a choice of 3, 4, 5, or 6 followed by a choice of h or t. There are 16 possible outcomes, but this sample set is not equally likely. The reason is that the rolls of 1 or 2 each show up with four choices that follow rather than two as is the case for 3, 4, 5, 6. You are just as likely to roll a 1 as a 3, but it does not appear this way in the sample set. Instead, suppose we have four choices that follow for 3, 4, 5, 6. Half of these choices we will read as h and half as t. This gives us the correct probability for heads if your roll a 3, 4, 5, or 6. In this new fake sample set, there are four choices for each of 3, 4, 5, 6 and two for each will result in us declaring that we have seen an h (for example, maybe we only look at the first toss). Now, there are $6\times 4$ elements in the fake sample set, four choices for each of the rolls. How many ways do you get an h? In the first part of the sample set, there are three ways for each of the rolls 1 and 2, so there are $2\times 3$ ways. In the second part, where the rolls are 3, 4, 5 or 6, there are two ways each (this is where things have been faked). We then have $4\times 2$ ways. This sums to 14 ways total in the fake sample set. \[\] As I said, this is not an ideal way of solving the problem, but it is very instructive. We'll go into more detail on Wednesday.