Simple Probability (1 Viewer)

[lyounamu]

The numbers 1,2,3,4, .... , 9 are written on each of 9 cards, 3 of the cards are chosen at random.

i) Find the probability that the sum of the three cards is equal to 9.

For this part, I just collected the combination of numbers that yield 9:
1,2,6
1,3,5
1,5,3
1,6,2
2,1,6
2,3,4
2,4,3
2,6,1
3,1,5
3,2,4
3,4,2
3,5,1
4,2,3
4,3,2
5,1,3
5,3,1

out of 10 . 9 . 8 = 720 combinations (i.e. 16/720 = 1/45)

I am not sure if I did it right.

2nd part) If it is known that first number chosen is 2, find the probability now that the sum of the 3 numbers is equal to 9.

Here, what I did was:

There are 4 combinations starting with 2 that give 9.

So 4 divided by (9.8) = 1/18.

Would this be right? Any suggestion will be a great help.

 

[lolokay]

your list isn't right. it has permutations, and even then not all of them - you forgot 6,2,1 and 6,1,2 which would give you 18

1,2,6
1,3,5
2,3,4

= 3 combinations (you can multiply by 3! to get the permutations)

total combinations = 9!/6!3! = 84

so probability is 3/84 = 1/28

for the second part, possible combinations of the next 2 numbers to give 9-2=7 are
1,6
3,5
= 2 combinations

out of a possible 8!/6!2! = 28

gives a probablity of 1/14

 

[shinn]

I think you have forgotten about the cases when 9 looks like 6 and vice versa in different orientations.