if 3 numbers are selected at random from a set X,

if 3 numbers are selected at random from a set X, such that Set X = {1,2,3,4,5,6,7}, what is the probability that the sum of the numbers selected is equal to 12? (Note: the same number can be selected again)

Since re-selection is allowed, total possibilities of selecting 3 numbers is 7^3.

For the sum of the selected numbers to be 12, they can be selected as:
7, 4, 1 --- 6 arrangements (7,4,1 or 7,1,4 or 4,1,7 or 4,7,1 or 1,7,4 or 1,4,7)
7, 3, 2 --- 6 arrangements
6, 5, 1 --- 6 arrangements
6, 4, 2 --- 6 arrangements
5, 4, 3 --- 6 arrangements
3, 3, 6 --- 3 arrangements (3,3,6 or 3,6,3 or 6,3,3)
2, 5, 5 --- 3 arrangements
and
4, 4, 4

total 37 arrangements or possibilities

Thus, answer is 37/7^3

The Answer is Option C

Selecting 3 digits from Set X = {1,2,3,4,5,6,7} with repetition allowed = n^r = 7^3

There are 37 ways 3 numbers can be selected to have a sum of 12.
1 + 5 + 6 = 3! = 6
1 + 4 + 7 = 3! = 6
2 + 3 + 7 = 3! = 6
2 + 4 + 6 = 3! = 6
2 + 5 + 5 = 3!/2! = 3
3 + 3 + 6 = 3!/2! = 3
3 + 4 + 5 = 3! = 6
4 + 4 + 4 = 3!/3! = 1
----------------------
Favorable cases - 37

Probability = Favorable cases / Total cases = 37 / 7^3.
Ans C

Favourable cases are
(1,4,7) (2,5,5) (2,4,6) (2,3,7)
(3,3,6) (3,4,5) (4,4,4)
Now
(1,4,7),(2,3,7), (2,4,6) & (3,4,5) each can be arranged in 3! =6 ways
So 4×6=24 cases
(2,2,5) &(3,3,6) each can be arranged in 3!/2!=3ways.
So 2×3=6cases
(4,4,4) is only 1 case
So total favourite cases are
24+6+1=31
All possible outcomes= 7^3
Hence required probability=31/7^3
Optin A is correct

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Ans : C 37/ 7^3

Looking for a short form to solve it...

Answer is \(37/(7^3)\)

Given: Set X = {1,2,3,4,5,6,7}
To find: Probability that the sum of the three numbers selected is equal to 12.


Step 1 : Find total number of ways 3 numbers can selected from Set X
For the first selection, we have 7 number to choose from. Therefore, number of ways = 7.
For second selection, we again have 7 numbers to choose from since number can be repeated. Therefore, number of ways is now = 7*7
For third selection, we again have 7 numbers to choose from since number can be repeated. Therefore, number of ways is now = 7*7*7

Therefore, number of ways 3 numbers can be selected from set X given that numbers can be repeated = \(7^3\)



Step 2 : Find the number of possibilities such that the sum of selected 3 numbers is 12.

Listing down all the possible combinations from Set X that form sum 12.

1) \(1+4+7\)
Number of arrangements possible for 1, 4 and 7 = \(3!\)
= \(6\)

2) \(1+5+6\)
Number of arrangements possible for 1, 5 and 6 = \(3!\)
= \(6\)

3) \(2+3+7\)
Number of arrangements possible for 2, 3 and 7 = \(3!\)
= \(6\)

4) \(2+4+6\)
Number of arrangements possible for 2, 4 and 6 = \(3!\)
= \(6\)

5) \(2+5+5\)
Number of arrangements possible for 2, 5 and 5 = \(3!/2!\)
= \(3\)

6) \(3+3+6\)
Number of arrangements possible for 3, 3 and 6 = \(3!/2!\)
=\( 3\)

7) \(3+4+5\)
Number of arrangements possible for 3, 4 and 5 = \(3!\)
= \(6\)

8) \(4+4+4\)
Number of arrangements possible for 4, 4 and 4 = \(3!/3!\)
= \(1\)

Total number of ways for sum to be 12 = \(6+6+6+6+3+3+6+1\)
= 37

Step 3 : Calculate the probability
Probability that sum of three number selected is 12 = Total number of ways for sum to be 12/Total number of ways 3 numbers can be selected
= \(37/7^3\)

given set x ; {1,2,3,4,5,6,7} ; target the probability that the sum of the numbers selected is equal to 12? (Note: the same number can be selected again)
least sum with repetition possible is
4,4,4 which can be arranged in 1 way
7 ; 7,4,1 & 7,3,2; 6 +6 :12ways
6 ; 6,5,1 ; 6,42, ; 6,33 ; possible ways ; ( 6+6+3) ; 15
5 ; 5,5,2 ; 5,4,3 ; possible ways ; 3+6 ; 9
total possible ways to get sum 12 with repetition ; 1+9+15+12 ; 37
7 digits can be arranged in 3 ways ; in 7^3 ways

37/(7^3) ; option C

if 3 numbers are selected at random from a set X, such that Set X = {1,2,3,4,5,6,7}, what is the probability that the sum of the numbers selected is equal to 12? (Note: the same number can be selected again)
A) 31/(7^3)
B) 35/(7^3)
C) 37/(7^3)
D) 35/7C3
E) 37/7C3

probability will be 35/(7^3)

Option B

We start by calculating 7C3 which equals 35.
Since this a probability question it's quite obvious that we cannot get a number >=1 so D and E are out.

Then we figure out a few of the possibilities:
1+5+6=12 3!=6
2+3+7=12 3!=6
and so on....this leads to 31 as the numerator.

Answer is A.

if 3 numbers are selected at random from a set X, such that Set X = {1,2,3,4,5,6,7}, what is the probability that the sum of the numbers selected is equal to 12? (Note: the same number can be selected again)

Soution:
No of ways sum is 12 with no numbers repeating, set of no (7-4-1), (7-3-2), (6-5-1), (5-4-2), (5-4-3)
each of five sets (7-4-1), (7-3-2), (6-5-1), (5-4-2), (5-4-3) can be selected in 6 ways e.g (7-4-1), (7-1-4), (4-7-1), (4-7-1), (1-7-4), (1-4-7) => no of ways= 5*6=30

No of ways sum is 12 with one number repeating, set of no (5-5-2), (3-3-6)
each of two sets (5-5-2), (3-3-6) can be selected in 3 ways e.g (5-5-2), (5-2-5), (2-5-5) => no of ways= 2*3=6

No of ways sum is 12 with all numbers repeating, set of no (4-4-4) => no of ways= 1[/color]

Total no of favorable outcomes = (30+6+1) =37

Total no of outcomes = 7*7*7 (as the same number can be selected again)[/color]

Probability = 37/ 7^3

Can anyone please explain why does the arrangement matter here?

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