What is the probability that the sum of the two numbers is 7?
[what i did]
0+7
1+6
2+5
4+3
my ans: 4/8
Is this right or wrong? Does without replacement mean we can't use the same number twice? IE: 4+2 AND 2+4?
Thanks for the help
$\begingroup$
$\endgroup$
Add a comment
|
3 Answers
$\begingroup$
$\endgroup$
3
Replacement does indeed mean that the same one cannot be used twice.
Order is not important, so we are using combinations in this case. Out of $_8 C_2 = 28$ possible sums, only 4 (7 + 0, 6 + 1, 5 + 2, 4 + 3) will yield 7. The probability is then $\frac{4}{28} \times 100\% = 14.3\%$.
-
1$\begingroup$ 8 C 2, since $\{ 0,1,2,3,4,5,6,7\}$ is 8 elements. $\endgroup$ Commented Jul 9, 2013 at 1:50
-
$\begingroup$ @Poseidonium why did you do 8C2? $\endgroup$ Commented Jul 9, 2013 at 2:33
-
1$\begingroup$ @MethodManX There are eight possible digits from which you can choose. You are choosing two of them. Order is not important, so we use combinations to remove the redundancies. $\endgroup$ Commented Jul 9, 2013 at 3:33
$\begingroup$
$\endgroup$
1
Because of typographical problems, it is not clear at this time what your answer is. So we write out an answer, actually two answers. Then you can decide whether you had the right solution/answer.
Without Replacement: Here by assumption the two numbers drawn are distinct. There are $\binom{8}{2}$ ways to choose $2$ numbers from the $8$ available. These are all equally likely.
You saw that there are $4$ pairs with sum $7$. So our probability is $\dfrac{4}{\binom{8}{2}}$.
If you calculate, you will see this simplifies to $1/7$.
With Replacement: Record the results as a pair $(x,y)$, where $x$ is the result of the first pick, and $y$ is the result of the second. There are $8^2$ such ordered pairs, all equally likely.
There are $8$ ordered pairs that give sum $7$. For note that as ordered pairs $(2,5)$ and $(5,2)$ are different. So our probability is $8/8^2$.
answered Jul 9, 2013 at 1:58
-
$\begingroup$ how do you find the 4 pairs without listing them out? Say you are choosing out of $\{1,2,3,...,30\}$ and the two randomly chosen integers need to have a sum of $30$? $\endgroup$ Commented Feb 22, 2020 at 13:57
$\begingroup$
$\endgroup$
Consider that no matter which first number you pick, there exists only one of the remaining 7 that will give you a total of 7 for the two numbers together. Thus, $\frac{1}{7}$ is the probability that would be the simplest way to see it from my perspective.
Without replacement means you can't pick the same number twice. For example, you couldn't choose 4 and then 4 again. This is what makes the problem a bit easier since that first number is taken out of the list and if you look at the list of values, they do pair up quite nicely to sum to 7 in each case.
