Problem Solving

Please time yourself...

A telephone number contains 10 digits ,including 3 digit area code . Bob remembers the area code and the next 5 digits of the number. He also remembers that the remaining numbers are not 0,1,2,5 or 7. If Bob tries to find the numbers by guessing the remaining digits at random,the probability that he will be able to find the correct number in at most 2 attempts is closest to which of the following?

A) 1/625
B) 2/625
C) 4/625
D) 25/625
E) 50/625

OA after some discussion.

My Ans - E

1:14
10 digits, 3 area code is known and next 5 digits known
thus unknown=2 digits
no. can be 2,3,6,8,9
total options=5*5=25
if in 1st attempt then 1/25
if in second, (24/25)*(1/24) = 1/25
total=2/25 = 50/625
IMO E

Total options = 25

Prob. correct in 1st attempt = 1/25
Prob. correct in 2nd attempt = 24/25 * 1/24 = 1/25

Prob. (at most 2 attempts) = 1/25 +1/25 = 2/25 = 50/625

total options=5*5=25

OA is E

I am not able to understand how total options=5*5=25 ?

This is the only thing which I could not think while solving the problem under 2 mins. Damn!! I am so mad at myself right now

smackmartine wrote:

total options=5*5=25

OA is E

I am not able to understand how total options=5*5=25 ?

This is the only thing which I could not think while solving the problem under 2 mins. Damn!! I am so mad at myself right now :(

oh I think I got it... two places to be filled in 5 and 5 ways because repetition is allowed. ahhh!

cans wrote:1:14
10 digits, 3 area code is known and next 5 digits known
thus unknown=2 digits
no. can be 2,3,6,8,9
total options=5*5=25
if in 1st attempt then 1/25
if in second, (24/25)*(1/24) = 1/25
total=2/25 = 50/625
IMO E

there are 5 unknown digits.In first attempt is the probability of getting it right not 1/5? I don't quite get this solution

baladon99 wrote:

cans wrote:1:14
10 digits, 3 area code is known and next 5 digits known
thus unknown=2 digits
no. can be 2,3,6,8,9
total options=5*5=25
if in 1st attempt then 1/25
if in second, (24/25)*(1/24) = 1/25
total=2/25 = 50/625
IMO E

there are 5 unknown digits.In first attempt is the probability of getting it right not 1/5? I don't quite get this solution

5 possible digits. But we have 2 spaces left (total 10, 3 of area code known and next 5 known. Thus remaining = 2 (=10-3-5))
let those be --
Now first space can be filled by 5 ways and next space can be filled by 5 ways. thus total of 5*5=25
Now correct answer is only one number (say either 89 or 99 or any one)
thus probability is 1/25 (one out of 25 numbers)

Cans!!! could you please explain me the following part in the above sum in more detail??

Prob. correct in 1st attempt = 1/25
Prob. correct in 2nd attempt = 24/25 * 1/24 = 1/25

[email protected] wrote:Cans!!! could you please explain me the following part in the above sum in more detail??

Prob. correct in 1st attempt = 1/25
Prob. correct in 2nd attempt = 24/25 * 1/24 = 1/25

we have total 5 digits with which we can form a 2 digit number.
Total no. of ways = 5*5 = 25 (-- for 1st space we have 5 options and for 2nd space, again we have 5 options)
But there is only one correct 2 digit number.
Thus prob correct in 1st attempt = 1/25 (1 correct and 25 possible)
prob correct in 2nd attempt means we got wrong in 1st and got correct in 2nd attempt
wrong in 1st attempt means (24/25) (24 wrong numbers and 25 total)
correct in 2nd attempt means (1/24) (1 correct and also out of 25, we used one number in 1st attempt and thus remaining=24)
thus prob in 2nd attempt = (24/25)(1/24)=1/25

Example: correct number is 89.
prob we get 89 in first attempt = 1/25
prob we get 89 in 2nd attempt means: we got wrong say 29 in first and then we got 89.
when we attempt again, we won't try 29 (as we know its wrong and thus 24 numbers left and that's why 1/24)

cans wrote:1:14
10 digits, 3 area code is known and next 5 digits known
thus unknown=2 digits
no. can be 2,3,6,8,9
total options=5*5=25
if in 1st attempt then 1/25
if in second, (24/25)*(1/24) = 1/25
total=2/25 = 50/625
IMO E

I understood till 1/25 for first attempt and 24/25 * 1/24 for second attempt.. I didnt get finally why are we adding them not multiplying them.. This is the confusion i have for many probems..

finites wrote:

cans wrote:1:14
10 digits, 3 area code is known and next 5 digits known
thus unknown=2 digits
no. can be 2,3,6,8,9
total options=5*5=25
if in 1st attempt then 1/25
if in second, (24/25)*(1/24) = 1/25
total=2/25 = 50/625
IMO E

I understood till 1/25 for first attempt and 24/25 * 1/24 for second attempt.. I didnt get finally why are we adding them not multiplying them.. This is the confusion i have for many probems..

We are talking about 2 scenarios independently i.e.
1) Successful attempt(1/25)
2) Failed Attempt * Successful Attempt (24/25 * 1/24)

We ADD the probabilities of each independent scenario to get the overall probability.
(As we did 1/25 + 1/25 )

If within a scenario, we need to calculate few steps, then we MULTIPLY the probabilities of each of those steps to get the probability of a scenario.
(As we did 24/25 * 1/24)

cans wrote:

[email protected] wrote:Cans!!! could you please explain me the following part in the above sum in more detail??

Prob. correct in 1st attempt = 1/25
Prob. correct in 2nd attempt = 24/25 * 1/24 = 1/25

we have total 5 digits with which we can form a 2 digit number.
Total no. of ways = 5*5 = 25 (-- for 1st space we have 5 options and for 2nd space, again we have 5 options)
But there is only one correct 2 digit number.
Thus prob correct in 1st attempt = 1/25 (1 correct and 25 possible)
prob correct in 2nd attempt means we got wrong in 1st and got correct in 2nd attempt
wrong in 1st attempt means (24/25) (24 wrong numbers and 25 total)
correct in 2nd attempt means (1/24) (1 correct and also out of 25, we used one number in 1st attempt and thus remaining=24)
thus prob in 2nd attempt = (24/25)(1/24)=1/25

Example: correct number is 89.
prob we get 89 in first attempt = 1/25
prob we get 89 in 2nd attempt means: we got wrong say 29 in first and then we got 89.
when we attempt again, we won't try 29 (as we know its wrong and thus 24 numbers left and that's why 1/24)

hello cans

i have some doubts same as amit....ur details elaboration makes it clear

regards nafi

Thank you Cans!! so much for such a wonderful explanation...

we know that bob has 5 digits to choose from. the probably of the right two-digits on the first attempt for 5 digits is
1/5.1/5(remember bob can repeat digits) = 1/25...as all the denominators are large numbers we will multiply this by 25 to get 25/625. as we can see that answer choices a,b, and c all are too small thus they are wrong. the question ask for the right answer in at least 2 attempts. the first attempt answer is 25/625 which is choice D, thus its also wrong as when we add the probability of getting right digits in 2nd attempt, our answer of 25/625 will increase.
now what if we get the wrong two number combination in first attempt
1-1/25 = 24/25
and getting the right answer in second attempt will equal 24/25. 1/25 (this fraction represent getting right answer in 2nd attempt = 24/625
adding all values
getting answer right in first attempt + getting answer wrong in first attempt but right in 2nd attempt = 25/625 + 24/625 = 49/625. the question ask that which fraction is close to the answer. 49/625 is close to 50/625. thus E is the right answer.