Some Help :) Probability (1 Viewer)

[wrxsti]

a credit card holder requires a 4 digit PIN number to withdraw her money... the digits can be 0,1,2,3,4,5,6,7,8,9

a) ifshe remembered that either the second OR the third digit was either a 6 or 7 what is the probability of guessing the PIN correctly


b) A member of the card holders family borrowed the card and he can remember the four digits but not theri order... WHat is the probability he guess the PIN correctly

 

[wrxsti]

Bump?

 

[watatank]

a) there's four cases to consider. the pin number can either be

- - 6 -
- 6 - -
- 7 - -
- - 7 -

there's a 1/10 chance of getting the blank number correct..since there are 3 blanks and one certainty in each case the probability of each case is the same

so

P = 4 * ( 1/10 * 1/10 * 1 * 1/10 )

= 4/1000

= 1/250

is this right?

 

[undalay]

b) just say 1,2,3,4 are the numbers.

i drew a tree diagram and found that when 1 was the first number there was 6 possible combinations. so it logically follows that when 2, 3 or 4 are first there will also be an extra 6 respectively. so 24 possible combinations in total. one of those *has* to be right, so the probability is 1/24.

hope that's right/that helps.

You didn't take into account that there must not be 4 unique digits. It could simply be 3 1's and 1 2, or something. Wouldn't you need to take that into consideration when calculating the probability?

 

[watatank]

yes its true i did assume they would be all different

then in that case...

(you might have to draw the tree diagrams to follow me)

case 1: all unique numbers, so just say 1,2,3,4 are the numbers.

i drew a tree diagram and found that when 1 was the first number there was 6 possible combinations. so it logically follows that when 2, 3 or 4 are first there will also be an extra 6 respectively. so 24 possible combinations in total.

case 2: two numbers are the same, so say the numbers are 1,1,2,3

when 1 is the first number there are 6 possible combos, again. then when the other '1' is the first number there are also 6, but they are the same 6 as when the other '1' was the first number. so it doesn't count.

when 2 is the first number there are 3 extra unique combinations, so it follows that 3 will also have 3 unique combinations.

so when there are two numbers that are the same there are 6 + 3 + 3 = 12 possible combinations

case 3: when there are three numbers that are the same, say 1,1,1,2 then 2 will just switch between spots so there are another 4 combinations

case 4: when all 4 are the same then there's 1 possible combination...

so all up there's 24 + 12 + 4 + 1 = 41 possible combinations...so 1/41 chance of getting the right one?

 

[iEdd]

a) There should be:
9x9x2x9 + 9x2x9x9 = 2916 combinations. Therefore her chance is 1/2916.

This is because each number can be one of 10 numbers (other than 6/7), except the 2nd digit, which can be 6 or 7, OR the third digit can be 6 or 7, so you add them. This is assuming she isn't sure if it's a six or a seven and which digit it belongs to, that is, no other digit can be that 6/7. The question is poorly worded as there are many ways to interpret it.

b) P = 1/4!
.: The probability of guessing it is 1/24
EDIT: As pointed out, that only works for unique digits.

 

[undalay]

yes its true i did assume they would be all different

then in that case...

(you might have to draw the tree diagrams to follow me)

case 1: all unique numbers, so just say 1,2,3,4 are the numbers.

i drew a tree diagram and found that when 1 was the first number there was 6 possible combinations. so it logically follows that when 2, 3 or 4 are first there will also be an extra 6 respectively. so 24 possible combinations in total.

case 2: two numbers are the same, so say the numbers are 1,1,2,3

when 1 is the first number there are 6 possible combos, again. then when the other '1' is the first number there are also 6, but they are the same 6 as when the other '1' was the first number. so it doesn't count.

when 2 is the first number there are 3 extra unique combinations, so it follows that 3 will also have 3 unique combinations.

so when there are two numbers that are the same there are 6 + 3 + 3 = 12 possible combinations

case 3: when there are three numbers that are the same, say 1,1,1,2 then 2 will just switch between spots so there are another 4 combinations

case 4: when all 4 are the same then there's 1 possible combination...

so all up there's 24 + 12 + 4 + 1 = 41 possible combinations...so 1/41 chance of getting the right one?

theres also 1122. But would you also need to take into account the possibility of that outcomes occuring? For example, for all digits to be the same, theres only 1 / 1000 of that occuring. Maybe i'm just overcomplicating the question :/

 

[wrxsti]

its a question 10a) of some trial paper...
i pretty much got to where you guys are. :S i got no answers to em though

 

[Ahmed A]

LOL
I just love the question you guys come across in your exams papers...

 

[wrxsti]

LOL
I just love the question you guys come across in your exams papers...

?????????????????????????????????????????????????????????????

 

[iEdd]

Nice.
The other thing that confused me (which is why mine was 9x9... and not 10x10..) was that if she thinks there is a 6/7 in there somewhere, does that mean she remembers the other digits to NOT be 6/7?