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I just want to know if my method is right:

P(Sum of 5 or At least one 4) = 2+3, 3+2, 4+1, 1+4 [+] (4+1,4+2,4+3,4+4,4+5,4+6)*2

So that will be 4+12/36

Ans: 16/36

am i right here?

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2 Answers 2

up vote 2 down vote accepted

Note that each of $(4, 1)$ and $(1,4)$ appear twice in each of the cases that you sum (those combinations are counted in the combinations that total $5$, and they are counted twice in the combinations in which at least one $4$ appears, so you're currently double counting each of those. You're also counting $(4, 4)$ twice, when doubling (to account for the permutation of) the combinations in which at least one $4$ appears. So you'd have to subtract a total of $3$ from your current total in the numerator:

$$\frac{4 + 12 - 3}{36} = \frac{[4 + (2\cdot 5 + 1)] - 2}{36} = \frac{13}{36}$$

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@ amwhy i only used 4,4 once can you please tell me what you mean by 4,4 twice? –  MethodManX Jul 5 '13 at 17:49
    
When you multiply by $2$: to account for counting the permutations of $(4, 1), (4, 2), (4, 3), {\bf (4, 4)}, (4, 5), (4, 6)$. When multiplying by $2$, you are counting the following as well: $(1, 4), (2, 4), (3, 4), {\bf (4, 4)}, (5, 4), (6, 4)$ –  amWhy Jul 5 '13 at 17:53
    
Doubling the first list, to account for these permutations, gives you $12$, but we need to throw out one of the counted $(4, 4)$, giving you $11$ pairs in which at least one $4$ appears. –  amWhy Jul 5 '13 at 17:59
    
@ amwhy okay thanks! –  MethodManX Jul 5 '13 at 18:04
    
You're welcome, MethodMan! –  amWhy Jul 5 '13 at 18:06

Making a list and carefully counting is a good idea. Even though it is not really necessary in this case, we will look at things in a more abstract way. Let $A$ be the event "sum is $5$" and let $B$ be the event "at least one $4$." Depending on whether you are in a counting mood or in a probability mood, we have $$|A\cup B| =|A|+|B|-|A\cap B|,\tag{1}$$ or $$\Pr(A\cup B) =\Pr(A)+\Pr(B)-\Pr(A\cap B).\tag{2}$$

In (1), we are using $|X|$ to denote the number of elemements in the set $X$. Your course may use a different notation.

The nice thing about the above formulas is that they can help organize the calculations.

We want to find $|A\cup B|$, and then divide by $36$ to find the probability. Or else we can use (2) directly. We will use (1), since it is more closely connnected with what you did.

First we find $|A|$, the number of ways we can have a sum of $5$. You made the count, it is $4$.

Next we find $|B|$, the number of ways to have at least one $4$. Note that there are $5^2$ ways to have no $4$, so there are $36-25=11$ ways to have at least one $4$. Or else we could list and count directly.

Next we find $|A\cap B|$. We want a sum of $5$ and at least one $4$. There are $2$ ways to do this.

Now from (1) we have $|A\cup B|=4+11-2$.

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