if this was in the GMAT. The level of this question would have be not more than 640.

Please note - here question is just asking you to find the probability not the number the ways in which Bob can dial the number.

1. He knows each digit could be one the five digits he remember, so probability in the first attempt would be (1/5*1/5) = 1/25.
2. Now the probability in the second attempt is something which i think would have tricked you,
consider carefully
--" let at the first attempt he guessed 1 and 0 for each digit (i.e. 1 out of 5 for each digit),
and at the second attempt he CAN choose 1 and 0 digits again, pairs might be different but he
can choose these two digits again.(it could be 1 and 5 or 3 or 0)'if you know what I mean' --"
so the probability in which the second trial could be made is 1/5*1/5 = 1/25
3. and now the final part, the total probability of two trials can not be multiplied as we do in the case of tossing a coin, for head or a tail, twice for a same out come.
i guess i can explain this too.
" consider tossing a coin twice for a same outcome, say a heads. coin tossed twice getting a head both the times, the probability would be (1/2*1/2=1/4) 'This part is same as two digits in the number, given in above example' , and second consider two coins tossing at the same time and after first toss, both the coins are tossed again, now the probability for getting a heads at first trial of tossing two coins together "OR" the second trial, again tossing the two coins together would be (1/4+1/4)"

so why we are not multiplying the 1/25 and 1/25 in the above example is it is "OR" in the question,
it is not "AND", as it would have been in the below question

"Bob and Brian remember 8 out of 10 digits in a number, but don't remember the remaining 2 digits. If they know that the remaining two digits are not 0,1,3,4,5 than what is the probability that both Bob AND Brian will dial or find the same number in a single attempt each is given?"

in this case it would have been 1/25*1/25


4. so finally answer to the above question would be 1/25+1/25 = 2/25


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This is just wrong!

There are 5*5 ways to pair the remaining digits for guessing the code.
The first trial is 1/25 or 1 way out 25 possible ways (pairs) gone.
The second trial is not tricky, as explained by previous poster, it's conditional probability. To perform the second trial we need to fail the first trial, i.e. Probability the second trial is success/Given Probability the first trial isn't success, (1-1/25)*1/24

Now sum the two probabilities to find the required, 1/25 + (24/25)*(1/24) = 2/25

[spoiler]2/25 is answer[/spoiler]

optimist.tageja wrote:if this was in the GMAT. The level of this question would have be not more than 640.

Please note - here question is just asking you to find the probability not the number the ways in which Bob can dial the number.

1. He knows each digit could be one the five digits he remember, so probability in the first attempt would be (1/5*1/5) = 1/25.
2. Now the probability in the second attempt is something which i think would have tricked you,
consider carefully
--" let at the first attempt he guessed 1 and 0 for each digit (i.e. 1 out of 5 for each digit),
and at the second attempt he CAN choose 1 and 0 digits again, pairs might be different but he
can choose these two digits again.(it could be 1 and 5 or 3 or 0)'if you know what I mean' --"
so the probability in which the second trial could be made is 1/5*1/5 = 1/25
3. and now the final part, the total probability of two trials can not be multiplied as we do in the case of tossing a coin, for head or a tail, twice for a same out come.
i guess i can explain this too.
" consider tossing a coin twice for a same outcome, say a heads. coin tossed twice getting a head both the times, the probability would be (1/2*1/2=1/4) 'This part is same as two digits in the number, given in above example' , and second consider two coins tossing at the same time and after first toss, both the coins are tossed again, now the probability for getting a heads at first trial of tossing two coins together "OR" the second trial, again tossing the two coins together would be (1/4+1/4)"

so why we are not multiplying the 1/25 and 1/25 in the above example is it is "OR" in the question,
it is not "AND", as it would have been in the below question

"Bob and Brian remember 8 out of 10 digits in a number, but don't remember the remaining 2 digits. If they know that the remaining two digits are not 0,1,3,4,5 than what is the probability that both Bob AND Brian will dial or find the same number in a single attempt each is given?"

in this case it would have been 1/25*1/25


4. so finally answer to the above question would be 1/25+1/25 = 2/25


-----------------------------------------------------

*Humble request
If you find this explanation satisfactory, please do click thanks button

Very well put pemdas, i appreciate your explanation.
but here my idea was to clear the doubt between an "AND" and "OR".

Thanks..

pemdas wrote:This is just wrong!

There are 5*5 ways to pair the remaining digits for guessing the code.
The first trial is 1/25 or 1 way out 25 possible ways (pairs) gone.
The second trial is not tricky, as explained by previous poster, it's conditional probability. To perform the second trial we need to fail the first trial, i.e. Probability the second trial is success/Given Probability the first trial isn't success, (1-1/25)*1/24

Now sum the two probabilities to find the required, 1/25 + (24/25)*(1/24) = 2/25

[spoiler]2/25 is answer[/spoiler]

optimist.tageja wrote:if this was in the GMAT. The level of this question would have be not more than 640.

Please note - here question is just asking you to find the probability not the number the ways in which Bob can dial the number.

1. He knows each digit could be one the five digits he remember, so probability in the first attempt would be (1/5*1/5) = 1/25.
2. Now the probability in the second attempt is something which i think would have tricked you,
consider carefully
--" let at the first attempt he guessed 1 and 0 for each digit (i.e. 1 out of 5 for each digit),
and at the second attempt he CAN choose 1 and 0 digits again, pairs might be different but he
can choose these two digits again.(it could be 1 and 5 or 3 or 0)'if you know what I mean' --"
so the probability in which the second trial could be made is 1/5*1/5 = 1/25
3. and now the final part, the total probability of two trials can not be multiplied as we do in the case of tossing a coin, for head or a tail, twice for a same out come.
i guess i can explain this too.
" consider tossing a coin twice for a same outcome, say a heads. coin tossed twice getting a head both the times, the probability would be (1/2*1/2=1/4) 'This part is same as two digits in the number, given in above example' , and second consider two coins tossing at the same time and after first toss, both the coins are tossed again, now the probability for getting a heads at first trial of tossing two coins together "OR" the second trial, again tossing the two coins together would be (1/4+1/4)"

so why we are not multiplying the 1/25 and 1/25 in the above example is it is "OR" in the question,
it is not "AND", as it would have been in the below question

"Bob and Brian remember 8 out of 10 digits in a number, but don't remember the remaining 2 digits. If they know that the remaining two digits are not 0,1,3,4,5 than what is the probability that both Bob AND Brian will dial or find the same number in a single attempt each is given?"

in this case it would have been 1/25*1/25


4. so finally answer to the above question would be 1/25+1/25 = 2/25


-----------------------------------------------------

*Humble request
If you find this explanation satisfactory, please do click thanks button