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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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02 Nov 2015, 10:15
1
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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02 Jul 2019, 09:38
srp wrote:
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

In case 1, when you attempted for the 1st time and got it wrong, 1 possibility out of 25 is exhausted. You are left with only 24 possibilities. So, the second attempts outcome should be measured on the remaining 24 possibilities only.

Then you will get the exact answer.
=(24/25)*(1/24)+(1/25)
=1/25 + 1/25
=2/25 which is same as 50/625.

Hope this helped.

Please correct me if I am wrong. Thank you.

Posted from my mobile device
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Re: A telephone number contains 10 digit, including a 3-digit  [#permalink]

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28 Jan 2020, 02:01
5 number in each of the two position total of 25,2 digit number.
Probability of getting it right in one guess = 1/25

Total probability= p(get it first try) + p( get it second try)
1/25+(25/25*1/25)

Posted from my mobile device
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Re: A telephone number contains 10 digit, including a 3-digit  [#permalink]

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28 Jan 2020, 22:11
incognito1 wrote:
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

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)

The answer is exactly E 50/625. With the rationale being
Getting it on the first try = 1/25
Getting it on the second try = 24/25*1/24 = 1/25

Hence total probability is 2/25 or 50/625.

In the case where he gets it the second try, the denominator should be 24 not 25 because he wouldn’t guess the same thing twice.
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Re: A telephone number contains 10 digit, including a 3-digit  [#permalink]

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21 Apr 2020, 13:28
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

We see that each of the remaining 2 digits is 3, 4, 6, 8 or 9.

The probability he can guess the remaining 2 digits correctly in the first attempt is 1/5 x 1/5 = 1/25.

The probability he can guess the remaining 2 digits correctly in the second attempt (provided that he guessed them incorrectly in the first attempt) is (1 - 1/25) x 1/25 = 24/25 x 1/25 = 24/625.

Therefore, the probability he can guess the remaining 2 digits correctly in at most 2 attempts is 1/25 + 24/625 = 25/625 + 24/625 = 49/625 ≈ 50/625.

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

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23 Apr 2020, 01:56
Permutation with repetition = N^r
5^2 = 25
There are 25 possible permutations

Probability of getting it right in 1 attempt = 1/25
Probability of getting it right in 2 attempts = 2/25

Answer = 2/25 = 50/625 = E
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A telephone number contains 10 digit, including a 3-digit  [#permalink]

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15 Jun 2020, 22:55
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 ?

What is the role of the phrase "AT MOST" here?

This problem is much easier if we ignore that part.
The answer is the same as the probability that he will find the correct number in 2 attempts anyway.

IMO, this 95% hard question is very very similar to this 25% hard question https://gmatclub.com/forum/sarah-cannot ... 40535.html
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Re: A telephone number contains 10 digit, including a 3-digit  [#permalink]

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16 Jun 2020, 06:36
1
varotkorn wrote:
What is the role of the phrase "AT MOST" here?

This problem is much easier if we ignore that part.

If you delete those words, then it's not clear what the question is asking. Do you want the probability he gets the phone number right precisely on his second attempt? Or in one of his first two attempts? Here we have 25 possibilities, one of which is right. If he'll pick two of the numbers, he'll have a 2/25 chance to pick the right one. Mathematically, the problem is essentially identical to a different question you asked about yesterday, which was something like this - if you have 25 employees, including Bob, and randomly pick a President and Vice-President, what is the probability Bob will be President or VP? The answer is 2/25.

Some solutions above produce the answer 49/625, which is not correct, unless he might pick the same phone number twice. You can see why that method is wrong just by imagining there are only two possible phone numbers. Then in two attempts Bob should have a 100% chance of getting the number right, but if you use the method that produces the answer 49/625, you would get an answer less than 100% to the two phone number question.

I have no idea why, in the original question, they appear to be asking for an estimate, nor why the fractions in the answer choices are not reduced.
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A telephone number contains 10 digit, including a 3-digit  [#permalink]

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16 Jun 2020, 06:47
IanStewart wrote:
varotkorn wrote:
What is the role of the phrase "AT MOST" here?

This problem is much easier if we ignore that part.

If you delete those words, then it's not clear what the question is asking. Do you want the probability he gets the phone number right precisely on his second attempt? Or in one of his first two attempts?

However, IMHO if the question were the probability that he will be able to find the correct number in at most 2 attemptS, then it's clear that there must be more than 1 attempts. So, it can't be his first or second attempt alone.

Moreover, as I learned from you yesterday, those 2 attempts can be from the tenth and eleventh attempts (not necessarily the first 2) and the result is still the same

IanStewart wrote:
Some solutions above produce the answer 49/625, which is not correct, unless he might pick the same phone number twice.

Totally agree here! Some imprecise reply above is actually of expert in quant
That's why I really need your view on this
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Re: A telephone number contains 10 digit, including a 3-digit  [#permalink]

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16 Jun 2020, 06:57
1
varotkorn wrote:
However, IMHO if the question were the probability that he will be able to find the correct number in at most 2 attemptS, then it's clear that there must be more than 1 attempts. So, it can't be his first or second attempt alone.

He presumably stops guessing when he finds the right number. So he won't make two attempts half the time, when he guesses right the first time. That's why the question says "at most two attempts", rather than "in two attempts".

Well-written math questions often use some language that might, at first glance, appear unnecessary, because sometimes those extra words ensure the question isn't open to two different legitimate interpretations. The wording they use here is necessary, so no one can interpret the question to mean "on precisely his second attempt".
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Re: A telephone number contains 10 digit, including a 3-digit   [#permalink] 16 Jun 2020, 06:57

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