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A telephone number contains 10 digit, including a 3-digit

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A telephone number contains 10 digit, including a 3-digit  [#permalink]

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New post Updated on: 02 Jul 2013, 11:02
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A telephone number contains 10 digit, including a 3-digit area code. Bob remembers the area code and the next 5 digits of the number. He also remembers that the remaining digits are not 0, 1, 2, 5, or 7. If Bob tries to find the number 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

Originally posted by marcodonzelli on 05 Feb 2008, 22:26.
Last edited by Bunuel on 02 Jul 2013, 11:02, edited 1 time in total.
Edited the question and added the OA.
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A telephone number contains 10 digit, including a 3-digit  [#permalink]

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New post 16 Jun 2015, 22:07
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elizaanne wrote:
The total number of possibilities for the phone numbers is 25,

Therefore the probability of him getting on the firs try is 1/25

Here's where I differ from other posters: I would say that the probability of him getting it right on the second try would be:

(24/25)(1/24)

This is because the probability of him getting it wrong on the first try is 24/24, because there are 24 wrong answers and 1 right one. After that, however, he's already eliminated one possible wrong answer by trying and failing, so the total number of possibilities is now 24. That means he has a 1/24 chance of getting it right after trying one and failing.

This makes the total probability 1/25+1/25, which is exactly 50/625


You are absolutely correct in your approach. However there seems to be a Typo error on the highlighted part above which should be 24/25


To Answer such question in just Two step you may consider taking Unfavorable case and calculate the favorable at second step

e.g. Probability of him not finding the correct Number in two steps = (24/25)*(23/24) = (23/25)

i.e. Probability of him finding the correct Number in two steps = 1- (23/25) = (2/25) which is same as 50/625
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Re: digits probability  [#permalink]

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New post 06 Feb 2008, 00:18
16
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I am getting E (as closest)

Remaining numbers to fill last two digits (3,4,6,8,9): Total 5

Probability of choosing right numbers in two places = 1/5 * 1/5 = 1/25
Probability of not choosing right numbers in two places = 1-1/25 = 24/25
--
At most two attempts: 1) Wrong-1st Attempt, Right - 2nd, 2) Right - 1st Attempt
1) = 24/25 * 1/25 = 24/625
2) = 1/25

Add 1 and 2, 24/625 + 1/25 = 49/625 ~ 50/625
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Re: digits probability  [#permalink]

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New post 04 Oct 2011, 15:03
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I think it is 1/25 (Correct in first attempt) + 24/25*1/24 (Correct in 2nd Attempt, it is 1/24 coz he wont repeat the wrong number again.) = 2/25 = 50/625 (Answer E)
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Re: digits probability  [#permalink]

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New post 12 Feb 2008, 10:01
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jackychamp wrote:
I have a question in this,
We have a pool of 5 numbers to choose from out of which we have to select 1.
So, prob of getting correct is 1/5.
which means probability of not correct is 4/5.
So, getting wrong at both places is 4/5 * 4/5 = 16/25.

Therefore getting right at both places is 1 - 16/25 = 9/25.

Can someone tell me where I am going wrong.

-Jack


You are missing out that you have to choose 2 correct numbers.

So, probabilty of 1st correct # is 1/5
probabilty of 2nd correct # is 1/5

So the probability of choosing correct combination is 1/5* 1/5 = 1/25.
Choosing wrong combination : 1 - 1/25 = 24/25

Now, we hv 2 cases.
case 1 : 1st attempt is wrong and second attempt is right.

24/25*1/25 = 24/625

case 2 : hit the eye in 1st attempt :
1/5

Total probability :

24/625 + 1/25 = 49/625 ~ 50/625
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Re: digits probability  [#permalink]

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New post 08 Feb 2008, 17:16
2
I have a question in this,
We have a pool of 5 numbers to choose from out of which we have to select 1.
So, prob of getting correct is 1/5.
which means probability of not correct is 4/5.
So, getting wrong at both places is 4/5 * 4/5 = 16/25.

Therefore getting right at both places is 1 - 16/25 = 9/25.

Can someone tell me where I am going wrong.

-Jack
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Re: digits probability  [#permalink]

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New post 10 Feb 2008, 11:40
2
1
jackychamp wrote:
I have a question in this,
We have a pool of 5 numbers to choose from out of which we have to select 1.
So, prob of getting correct is 1/5.
which means probability of not correct is 4/5.
So, getting wrong at both places is 4/5 * 4/5 = 16/25.

Therefore getting right at both places is 1 - 16/25 = 9/25.

Can someone tell me where I am going wrong.

-Jack


Good question!

[Probability (getting both digits right)] is NOT the same as [1 - Probability (getting both digits wrong)]
Since if either digit is wrong, the guessed number is wrong.

[Probability (getting both digits right)] = 1 - [Probability (getting both digits wrong) + Probability (getting either digit wrong)]
= 1 - [4/5 + (4/5 * 1/5) + (1/5 * 4/5)]
= 1/25

Now you have two tries to guess the correct number.
Probability (getting it right the first time) = 1/25
Probability (getting it wrong the first time, and right the second time) = (1 - 1/25) * 1/25 = 24/625

Add the two together, you get 49/625, closest to (E)
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Re: digits probability  [#permalink]

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New post 10 Feb 2008, 12:27
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marcodonzelli wrote:
A telephone number contains 10 digit, including a 3-digit area code. Bob remembers the area coe and the next 5 digits of the number. He also remembers that the remaining digits are not 0,1,2,5, or 7. If Bob tries to find the number 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 ?

1/625

2/625

4/625

25/625

50/625


its E. 50/625 = 2/25

the telephone number is: xxx-xxxxx-ab
we need to find the digits ab:
0, 1, 2, 5, and 7 cannot be in digits ab.
remaining integers 3, 4, 6, 8, and 9 can be chosen for ab.
no of ways the integers can be put in place of ab = 5x5 = 25

prob = 1/25 + 1/25 = 2/25. which is E.
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Re: A telephone number contains 10 digit, including a 3-digit  [#permalink]

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New post 02 Jul 2013, 10:56
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remaining digits are 3.4.6.8.9

ways to choose last two digits = 5*5 = 25

P(choosing correct number in 1st attempt) = 1/25

P(choosing wrong in 1st attempt) = 24/25
P(choosing right in 2nd attempt) = 1/24 {i am assuming he will not try wrong number again}

therefore total probability = 1/25 + 24/25*1/25 = 2/25 = 50/625 (E).... i don't think this question requires rounding off.... Correct me if i missed any step
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Re: A telephone number contains 10 digit, including a 3-digit  [#permalink]

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New post 07 Aug 2013, 08:02
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I tried to solve this question in following manner:

Last 2 digit can be chosen out of 3 4 6 8 9
means 5x5 = 25 Numbers can be constructed.

Case 1: Bob able to dial the number in first attempt : 1/25
Case 2: Bob able to dial the number in second attempt :

24/25 x 1/24 = 1/25

case 1 + case 2 = 2/25 which is same as 50/625.

I have seen someone solved second case as 24/25 x 1/25 which is not right.
bcz in second attempt bob will have only 24 choices not 25.
thats why they got 49/625 or approx answer.
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Re: A telephone number contains 10 digit, including a 3-digit  [#permalink]

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New post 17 May 2014, 08:05
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JusTLucK04 wrote:
I guess he is hitting randomly here and the probability that he chooses a number at any point of time is equal..
1/25+24/25*1/25= 49/625
It should be explicitly stated if he is trying numbers in an order/not repeating...I think


I think most of the answers are missing a point. Let me try to put it across:

Total number of possible numbers are : 5x5 = 25

Correct number =1

Case 1: When he gets it right in first attempt: P(E1) = 1/25

Case 2: He gets 1st attempt wrong and second right:

When he gets it wrong then the probability of getting wrong is 24/25.
Now there are 24 cases with him and he chooses the right one this time.
Probability of right case is 1/24

Thus, P(E2) = 24/25 x 1/24

=1/25

Probability of getting it right in at most two cases = P(E1) + P(E2)
= 1/25 + 1/25
= 2/25
= 50/625
Option (E) is therefore right as most of you mentioned but the method employed was wrong.

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Re: A telephone number contains 10 digit, including a 3-digit  [#permalink]

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New post 16 Jun 2015, 21:25
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The total number of possibilities for the phone numbers is 25,

Therefore the probability of him getting on the firs try is 1/25

Here's where I differ from other posters: I would say that the probability of him getting it right on the second try would be:

(24/25)(1/24)

This is because the probability of him getting it wrong on the first try is 24/24, because there are 24 wrong answers and 1 right one. After that, however, he's already eliminated one possible wrong answer by trying and failing, so the total number of possibilities is now 24. That means he has a 1/24 chance of getting it right after trying one and failing.

This makes the total probability 1/25+1/25, which is exactly 50/625
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Re: digits probability  [#permalink]

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New post 06 Feb 2008, 01:02
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3,4,6,8,9 are the digits to be arranged in two positions X Y

total number of possible ways = 5*5 = 25

Case A:

X( ok) Y( ok ): (1/5)*(1/5) = 1/25


Case B:

First attempt

X(not ok) Y (not ok): (1- 1/25) = 24/25

Second attempt

X( ok) Y( ok ): (1/5)*(1/5) = 1/25


= 24/625

Case C:

First attempt

X( ok) Y(not ok) : (1/5)*(1- 1/5) = 4/25

Second attempt

X( ok) Y( ok ): (1/5)*(1/5) = 1/25

=4/625

Case D:

First attempt

X(not ok) Y(ok) : (1/5)*(1- 1/5) = 4/25

Second attempt

X( ok) Y( ok ): (1/5)*(1/5) = 1/25

=4/625



Total = (1/25) + ( 24/625 ) + ( 4/625 ) + ( 4/625 )

=

I think I am missing some case!
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Re: digits probability  [#permalink]

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New post 06 Feb 2008, 01:12
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marcodonzelli wrote:
A telephone number contains 10 digit, including a 3-digit area code. Bob remembers the area coe and the next 5 digits of the number. He also remembers that the remaining digits are not 0,1,2,5, or 7. If Bob tries to find the number 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 ?

1/625

2/625

4/625

25/625

50/625


Answer should be (E)

5 numbers in each of the last two positions
==> total of 25 2-digit numbers
==> probability of getting it right in one guess = 1/25

Total Probability = P(get it right the first try) + P(get it right the second try)
= 1/25 + [24/25 * 1/25]
= 49/625, closest to (E)
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Re: digits probability  [#permalink]

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New post 06 Feb 2008, 01:32
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marcodonzelli wrote:
A telephone number contains 10 digit, including a 3-digit area code. Bob remembers the area coe and the next 5 digits of the number. He also remembers that the remaining digits are not 0,1,2,5, or 7. If Bob tries to find the number 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 ?

1/625

2/625

4/625

25/625

50/625


The prob of getting it right in one attempt is 1/5*1/5 = 1/25

This is already bigger than ABC so elim. Also bigger than D so elim.

E is left.

You can do it this way as well 1/5*1/5 -> 1/25 and thus 24/25 is not getting it

1/25+24/25(1/25) --> 49/625
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A telephone number contains 10 digit, including a 3-digit  [#permalink]

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New post 02 Nov 2015, 11:15
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so Bob is given 2 opportunities to decipher the remaining 2 digits of the code, for this task he has 5 digits to choose from.

IMPORTANT notes to bear in mind before calculation:
- picking a number for place 9 is independent from picking the number for place 10 - i.e. digits can repeat
- if Bob wastes his 1st opportunity he will have better chances to pick the right combination during his second attempt - because by that time he has already proved that one combination is invalid.

So we have all the info needed to do simple calculation:

(1) \(\frac{1}{5}\)*\(\frac{1}{5}\) = \(\frac{1}{25}\) - this is the independent event that he would win on first attempt

OR

(2.1) \(\frac{4}{5}\)*\(\frac{4}{5} = [m]\frac{16}{25}\) - he fails to choose correctly on first attempt because he picks either of the 4 wrong digits for the both places

(2.2) \(\frac{1}{4}\)*\(\frac{1}{4}\)=\(\frac{1}{16}\) - he finally chooses the correct digit out of the remaining 4 for each place independently!

AND

(2.3) Multiply the above two iterations to complete the chances for the second scenario: \(\frac{16}{25}\)*\(\frac{1}{16}\)=\(\frac{1}{25}\)

(3) Sum up \(\frac{1}{25}\)+\(\frac{1}{25}\)=\(\frac{2}{25}\)=\(\frac{50}{625}\)
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Re: A telephone number contains 10 digit, including a 3-digit  [#permalink]

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New post 07 Aug 2013, 09:14
marcodonzelli wrote:
A telephone number contains 10 digit, including a 3-digit area code. Bob remembers the area code and the next 5 digits of the number. He also remembers that the remaining digits are not 0, 1, 2, 5, or 7. If Bob tries to find the number 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


solution:

Digits are available = 3,4,6,8,9
so two spaces can be filled by 5 numbers = 5 * 5 = 25 ways
chances of failing = 24
At most 2 attempts = success at 1st attempt or success at 2nd attempt.
= 1/25 + 24/25 * 1/25 = 49/625 =~ 50/625 (E) Answer
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Re: A telephone number contains 10 digit, including a 3-digit  [#permalink]

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New post 05 Feb 2014, 15:17
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marcodonzelli wrote:
A telephone number contains 10 digit, including a 3-digit area code. Bob remembers the area code and the next 5 digits of the number. He also remembers that the remaining digits are not 0, 1, 2, 5, or 7. If Bob tries to find the number 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


Good question +1

Probability of at least in 2 attempts = Prob in 1 attempt + Prob of Not 1 atempt * Prob of 2nd attempt

Therefore probability of 1 attempt = 1/25

Now this is where the problem gets tricky

Prob of not 1 could be

He found 1
OR He found the other
OR He found neither

So we have 3 scenarios that we'll need to add up

1st scenario: 1/5^3 * 4/5
2nd scenario: 1/5^3 * 4/5
3rd scenario: 4/5^2 * 1/5^2

Now adding all the scenarios = 24/625

Adding 1st and 2nd grand scenario = 1/25 + 24 /625 = 49 / 625.

The closest answer choice is E, cause the problem says approximately

Hope its clear
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Re: A telephone number contains 10 digit, including a 3-digit  [#permalink]

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New post 13 May 2014, 14:21
I guess he is hitting randomly here and the probability that he chooses a number at any point of time is equal..
1/25+24/25*1/25= 49/625
It should be explicitly stated if he is trying numbers in an order/not repeating...I think
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Re: A telephone number contains 10 digit, including a 3-digit  [#permalink]

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New post 18 May 2014, 04:55
mittalg wrote:
JusTLucK04 wrote:
I guess he is hitting randomly here and the probability that he chooses a number at any point of time is equal..
1/25+24/25*1/25= 49/625
It should be explicitly stated if he is trying numbers in an order/not repeating...I think


I think most of the answers are missing a point. Let me try to put it across:

Total number of possible numbers are : 5x5 = 25

Correct number =1

Case 1: When he gets it right in first attempt: P(E1) = 1/25

Case 2: He gets 1st attempt wrong and second right:

When he gets it wrong then the probability of getting wrong is 24/25.
Now there are 24 cases with him and he chooses the right one this time.
Probability of right case is 1/24

Thus, P(E2) = 24/25 x 1/24

=1/25

Probability of getting it right in at most two cases = P(E1) + P(E2)
= 1/25 + 1/25
= 2/25
= 50/625
Option (E) is therefore right as most of you mentioned but the method employed was wrong.

Cheers!!!


Whats wrong with my method..Can you please elaborate?..Wasn't able to catch you here

I mean to imply that he is not taking a note of the numbers hit earlier and is just going about randomly..as nothing has been mentioned
Anyways this is trivial and GMAC will be more specific..Skip it
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Re: A telephone number contains 10 digit, including a 3-digit   [#permalink] 18 May 2014, 04:55

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