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Two fair coins are to be tossed once. For each head that results, one fair die is to be rolled. What is the probability that the sum of the die rolls is odd? (Note that if no die is rolled, the sum is $0$.)

I thought that since odd+even =odd, we then have to get (odd,even) and (even,odd) pairs from $\{0,1,2,3,4,5,6\}$. Thus, since there are a total of $7*7$ possibilities for pairs and $4*3*2$ pairs that add to an odd number I get $\dfrac{4*3*2}{7*7} = \dfrac{24}{49}$ but the right answer is $\dfrac{3}{8}$. What did I do wrong?

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    $\begingroup$ tbh I don't quite understand where that fraction is coming from. $\endgroup$ – joedoe8585 Jan 23 '16 at 19:00
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There is a $1/4$ chance of no heads, and therefore no dice. The sum of no dice is $0$, which is even.

There is a $1/2$ chance of one head, and therefore one die. For one die, there is a $1/2$ chance of landing on an odd total $1,3,5$.

There is a $1/4$ chance of getting two heads, and therefore two dice. There are an equal number of possibilities giving even and odd, so there is again a $1/2$ chance of landing on an odd total.

So in total, we have total probability $$ \frac{1}{4} \cdot 0 + \frac{1}{2} \cdot \frac{1}{2} + \frac{1}{4} \cdot \frac{1}{2} = \frac{3}{8},$$ as you indicated was the correct answer. $\diamondsuit$

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  • $\begingroup$ I was asking what I did wrong in my approach not what the correct approach was. $\endgroup$ – user19405892 Jan 23 '16 at 19:15
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What you are doing wrong is assuming that each of those possible outcomes is equally likely.

Tossing the two coins there are four possible outcomes: Head, Head; Head, Tail; Tail, Head; Tail, Tail, each with probability 1/4.

In case of the outcome "Head, Head", two die are rolled. In this case, the outcomes and their probabilities are 2: prob 1/36; 3: prob 2/36= 1/18; 4: prob 3/36= 1/12; 5: prob 4/36= 1/9; 6: prob 5/36; 7: prob 6/36= 1/6; 8: prob 5/36; 9: prob 4/36= 1/9; 10: prob 3/36= 1/12; 11: prob 2/36= 1/18; 12: prob 1/36; The probability of "odd" is 1/18+ 1/9+ 1/6+ 1/9+ 1/18= 9/18= 1/2. Since the probability of this case is 1/4, multiply that by 1/4: 1.8.

In the case of "Heads, Tails" or "Tails, Heads" we roll a single die so the probability of any one number is 1/6. There are 3 odd numbers so the probability of "odd" is 3/6= 1/2 also. The probability of this case is 1/2 so multiply by 1/2: 1/4.

in the case of "Tails, Tails" we do not roll any die so the "sum" is 0. That is even so the probability of "odd" in this case is 0.

The overall probability of "odd" is 1/8+ 1/4= 3/8.

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Notice that not your sample set cannot be one of $\{0, 1, 2, 3, 4, 5, 6\}$ because they do not all occur with equal probabilities. Specifically, the set $\{1, 2, 3, 4, 5, 6\}$ only occurs with probability $\frac{1}{2}$ for each die (you might roll $T$ and not have those outcomes at all). Similarly, $\{0\}$ occurs with probability $\frac{1}{2}.$

Adjusting your calculation to account for this, we find that there is a $(\frac{1}{2} + \frac{1}{2}(\frac{1}{2})) \cdot (\frac{1}{2}(\frac{1}{2})) = \frac{3}{16}$ chance to get (even, odd) and another $\frac{3}{16}$ chance to get (odd, even) by symmetry. Our final answer is $2 \cdot \frac{3}{16} = \boxed{\frac{3}{8}}.$

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  • $\begingroup$ We have HH,TH,HT, and TT. For HH we have $\{1,2,\ldots,6\} \times \{1,2,\ldots,6\}$; for $TH$ we have $\{1,2,\ldots,6\} \times {0}$; for $HT$ we have $\{0\} \times \{1,2,\ldots,6\}$; and for TT we have $\{0\} \times \{0\}$. How are not these equally likely? $\endgroup$ – user19405892 Jan 23 '16 at 19:34
  • $\begingroup$ Look closer: is the probability of getting a $0$ the same as the probability of getting a $6?$ Of course not! @user19405892 $\endgroup$ – K. Jiang Jan 23 '16 at 19:42

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