Probability

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kavitha

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Dec 23, 2010
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Question:1
You select 2 numbers at random from 1,2,3,4,5,6,7,8. What is the probability that the sum of two numbers is 3?

I tried as follows:
1+2=3
p(1)=1/8
p(2)=1/8

1/8 * 1/8= 1/64

correct me if I am wrong.



Question 2:
You select 3 numbers at random from 1,2,3,4,5,6,7,8,9. What is the probability that the sum of 3 numbers is 6?

I tried as follows:
1+2+3=6
or
2+1+3=6
or
3+2+1=6


p(1)=1/9
p(2)=1/9
p(3)=1/9
1/9 * 1/9 *1/9 = 1/729



correct me if I am wrong.
 
kavitha said:
Question:1
You select 2 numbers at random from 1,2,3,4,5,6,7,8. What is the probability that the sum of two numbers is 3?

I tried as follows:
1+2=3
p(1)=1/8
p(2)=1/8

1/8 * 1/8= 1/64
correct me if I am wrong.




Question 2:
You select 3 numbers at random from 1,2,3,4,5,6,7,8,9. What is the probability that the sum of 3 numbers is 6?

I tried as follows:
1+2+3=6
or
2+1+3=6
or
3+2+1=6


p(1)=1/9
p(2)=1/9
p(3)=1/9
1/9 * 1/9 *1/9 = 1/729



correct me if I am wrong.
Click to expand...

Are you selecting one number then the other (With or without replacement) - or together?

Are you allowed to choose repeating numbers ( i.e. are you allowed to have 1+1+4 = 6)?
 
Yes, we are allowed to choose repetitive numbers.
Can be with replacement also.

Please guide me. I am new to this topic.
 
kavitha said:
Question:1
You select 2 numbers at random from 1,2,3,4,5,6,7,8. What is the probability that the sum of two numbers is 3?
I tried as follows:
1+2=3
p(1)=1/8
p(2)=1/8
1/8 * 1/8= 1/64
Click to expand...
No. p(1) = 1/4 ; 1 or 2 both acceptable: get it?
 
kavitha said:
Question:1
You select 2 numbers at random from 1,2,3,4,5,6,7,8. What is the probability that the sum of two numbers is 3?

I tried as follows:
1+2=3
p(1)=1/8
p(2)=1/8

1/8 * 1/8= 1/64

correct me if I am wrong.

It is 1/7.
Question 2:
You select 3 numbers at random from 1,2,3,4,5,6,7,8,9. What is the probability that the sum of 3 numbers is 6?

I tried as follows:
1+2+3=6
or
2+1+3=6
or
3+2+1=6


p(1)=1/9
p(2)=1/9
p(3)=1/9
1/9 * 1/9 *1/9 = 1/729



correct me if I am wrong.
Click to expand...
It is 2/7.
 
Question:1
You select 2 numbers at random from 1,2,3,4,5,6,7,8. What is the probability that the sum of two numbers is 3?

solution:
you are selecting 2 elements from (1 to 8)= 8C2 possibilities are there.

among that the condition given is sum is 3, you know that only one pair (1,2)

so, total no. of possibilities=8C2

required possibility=1

probability=1/8C2

i.e, p=1/28
 
kavitha said:
Can be with replacement also.
Click to expand...
That indicates that your problem has 2 solutions:
1 / 28 if no replacement
1 / 32 if replacement
 
hi denis,

selection and combination are synonymous in probability.

from 8 no. selecting 2 two no. is same as combination of 2 from 8 no. =8C2.

i think we should not confuse with replacement or without replacement.
 
saravananbs said:
from 8 no. selecting 2 two no. is same as combination of 2 from 8 no. =8C2.
Click to expand...
Agree. What are you trying to tell me?

The original problem is:
Question:1
You select 2 numbers at random from 1,2,3,4,5,6,7,8. What is the probability that the sum of two numbers is 3?

From a subsequent post by Kavitha, I understood (perhaps erroneously) that 2 questions were meant:
also with replacement, meaning same number could be picked twice.
Anyhow, I'm not here to argue...

EDIT: by the way, our esteemed Mister lookagain did show 1/32 as probability WITH REPLACEMENT;
but he has chosen to delete his post.
I'm looking at it this way:
8 balls numbered 1 to 8 are in a "spinner!".
1: using both hands at same time, you pick a ball in each hand; what is the probability that their sum will be 3? : 1/28

2: using one hand, you pick ONE ball; then you throw that ball back in the spinner;
now using one hand again, you pick another ball: what is the probability that the sum of the 2 balls is 3? : 1/32

2nd EDIT: good news!! I see that Soroban has added "with replacement" to original problem...
 
Hello, kavitha!

You select 2 numbers at random from {1,2,3,4,5,6,7,8} with replacement.
What is the probability that the sum of two numbers is 3?
Click to expand...

There are \(\displaystyle 8^2 = 64\) possible outcomes.

. . . . \(\displaystyle \boxed{\begin{array}{cccccccc} (1,1) & (1,2) & (1,3) & \hdots & (1,8) \\ (2,1) & (2,2) & (2,3) & \hdots & (2,8) \\ (3,1) & (3,2) & (3,3) & \hdots & (3,8) \\ \vdots & \vdots & \vdots & & \vdots \\ (8,1) & (8,2) & (8,3) & \hdots & (8,8) \end{array} }\)

Only two have a sum of 3: .\(\displaystyle (1,2),\:(2,1)\)


\(\displaystyle \text{Therefore: }\:p(\text{sum of 3}) \:=\:\frac{2}{64} \:=\:\frac{1}{32}\)



You select 3 numbers at random from {1,2,3,4,5,6,7,8,9} with replacement.
What is the probability that the sum of 3 numbers is 6?
Click to expand...

\(\displaystyle \text{There are: }\:6^3 = 216\text{ possible outcomes.}\)

\(\displaystyle \text{We have: }\;\begin{Bmatrix} (1,1,4) & \text{which has 3 permutations} \\ (1,2,3) & \text{which has 6 permutations} \\ (2,2,2) & \text{which has 1 permutation} \end{Bmatrix}\)

\(\displaystyle \text{Hence, there are: }\:3 + 6 + 1 \:=\:10\text{ outcomes with a sum of 6.}\)


\(\displaystyle \text{Therefore: }\:p(\text{sum of 6}) \;=\;\frac{10}{216} \;=\;\frac{5}{108}\)

 
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