Probability Question (1 Viewer)

[Skeptyks]

Hey just need help on a probability question that popped up in a practice MX1 paper.

In a gambling game, a computer will take, at random, thirteen numbers from the integers from 1 to 40 inclusive. Before this occurs, you are invited to choose two numbers, hoping the computer will 'match' them among the thirteen.

i) Show that the probability of having both your numbers ' matched ' is 0.1
ii) Find the probability that the computer will match neither of your numbers

If anyone could provide a detailed explanation on how to obtain the answer it would be great since I have the 'working out' already.

Thanks.

 

[deswa1]

The method that instantly jumps out at me is find the complementary event and subtract from 1. From part i), you worked out the chance of matching both numbers. Now find the chance that it will match one of your numbers. Add these two and subtract from 1 and there's your answer.

 

[Skeptyks]

So the answers say this:

Number of pairs you can choose = 40C2
Number of successful pairs you can choose = 13C2

Probability = 13C2/40C2 = 0.1

1) Probability = 13C2/40C2 = 0.45

Now I tried 1- (13C1/40C2 x 27C1/40C2 + 13C2/40C2) and it didn't work out very well.

If you could explain how the given solutions are done it would be better than correcting my mistake above.

Thanks.

 

[Skeptyks]

Anyone?

 

[Aesytic]

i did it differently to the answers, but got the same values
for part i, if the computer chooses both your numbers, then the number of combinations where this can happen are 2C2 * 38C11, since 2 of the 13 numbers chosen must be the numbers you chose, and then the computer chooses the other 11 numbers from the remaining 38. the total sample space is just the number of combinations where the computer chooses 13 numbers from 40, which is 40C13
.'. probability = (2C2*38C11)/40C13 = 0.1

for part ii, by using a similar method:
if the computer chooses neither of your numbers, then the total combinations of numbers it can have are 38C13, since 2 of the 40 numbers aren't chosen (the two that you chose). dividing by the sample space, which is 40C13 again, you get 38C13/40C13 = 0.45

 

[Skeptyks]

i did it differently to the answers, but got the same values
for part i, if the computer chooses both your numbers, then the number of combinations where this can happen are 2C2 * 38C11, since 2 of the 13 numbers chosen must be the numbers you chose, and then the computer chooses the other 11 numbers from the remaining 38. the total sample space is just the number of combinations where the computer chooses 13 numbers from 40, which is 40C13
.'. probability = (2C2*38C11)/40C13 = 0.1

for part ii, by using a similar method:
if the computer chooses neither of your numbers, then the total combinations of numbers it can have are 38C13, since 2 of the 40 numbers aren't chosen (the two that you chose). dividing by the sample space, which is 40C13 again, you get 38C13/40C13 = 0.45

Thanks this looks to me like a more logical method.