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by abhasjha » Thu Oct 16, 2014 9:33 pm
I want to know how can we approach questions of this type ?

What is the probablity of selecting a 6 digit number such that the fourth digit is the same as first digit , second digit is the same as fifth digit and third digit is the same as 6th digit ?

do not have the answer options available ....

can someone please tell me how to proceed with this ?
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by [email protected] » Thu Oct 16, 2014 10:03 pm
Hi abhasjha,

These types of questions are always carefully written, so you have to pay attention to the details and you have to think about the limitations for each digit.

For example, a 6-digit number won't start with a 0, but the other digits could be 0 (unless it's a numerical code, then it MIGHT be able to start with a 0). Here, I'll assume that the intent is to ask for a 6-digit number with no special rules to consider besides the ones that you listed....

The possible 6-digit numbers are 100,000 - 999,999, inclusive, so there are 900,000 possible numbers.

We're asked for the probability of finding a very specific type of 6-digit number

1st Digit: 1-9 = 9 possibilities
4th Digit: MUST match the 1st = 1 possibility (once we know the first digit)
2nd Digit: 0-9 = 10 possibilities
5th Digit: MUST match the 2nd = 1 possibility (once we know the second digit)
3rd Digit: 0-9 = 10 possibilities
6th Digit: MUST match the 3rd = 1 possibility (once we know the third digit)

So....(9)(10)(10)(1)(1)(1) = 900 possibilities that match the description

900/900,000 = 1/1,000 = the probability of finding the 6-digit number described.

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by Mathsbuddy » Fri Oct 17, 2014 1:10 am
Although the fist digit cannot be zero, the rest can.
However, the 1st, 2nd and 3rd digits are irrelevant, as we only need the probability of a match to each:

P(A) = P(1st = 4th) = 1/10
P(B) = P(2nd = 5th) = 1/10
P(C) = P(3rd = 6th) = 1/10

P(A AND B AND C) = 1/10 * 1/10 * 1/10 = 1/1000
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by GMATGuruNY » Fri Oct 17, 2014 1:49 am
abhasjha wrote:I want to know how can we approach questions of this type ?

What is the probablity of selecting a 6 digit number such that the fourth digit is the same as first digit , second digit is the same as fifth digit and third digit is the same as 6th digit ?

do not have the answer options available ....

can someone please tell me how to proceed with this ?
P(4th digit matches the 1st) = 1/10. (Of the 10 digits, only one will match the 1st.)
P(5th digit matches the 2nd) = 1/10. (Of the 10 digits, only one will match the 2nd.)
P(6th digit matches the 3rd) = 1/10. (Of the 10 digits, only one will match the 3rd.)
Since we want all of these probabilities to happen, we MULTIPLY the fractions:
1/10 * 1/10 * 1/10 = 1/1000.
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