A number is a palindrome if it reads exactly the same from right to left as it does from left to right. For example. 959 and 24742 are palindromes.

Quantity A

The probability chat a 3-digit number chosen at random is a palindrome
Qty B = 1/10
To me the answer was D

In Qty A there is no mention of what the denominator is: that is how many 3 digit palindromes out of 100 - 999 (only 3 digits) or 100 - 9999 (4 digit numbers). So the probability would differ.

Can someone explain why D is wrong?

Nice one. The important part is to figure out how many palindromes are there in those 900 three digit numbers.

Quantity A, without doubt

x-Y-x is a palindrom. Since 0-#-# would actually be a 2 digit number, we start with 1-#-#. For a 3 digit number, a palindom would be 1-(0-9)-1, 2-(0-9)-2, 3-(0-9)-3, and so on through 9-(0-9)-9. (0-9) is 0, 1,2,3,4,5,6,7,8,9 ...this is 10 possible answers. Therefore, there are 10 possible answers for 1-#-1, 10 possible for 2-#-2, 10 for 3-#-3 and so on. So now you know that there are 9 possible palindrom start/end digits that have 10 possible answers each. You can think of this as 10+10+10+10+10+10+10+10+10 or 10*9, which equals 90. This is the chance of a random 3 digit numbe being a palindrom.

Next we need to figure out how many possible numbers between 100-999 there are. The answer is 900. You can think of this as "there are 100 numbers between 0-99 and 1000 numbers between 0-999, then subtracting 100 from 1000 since we starts at 100 (skipping 0-99)."

Using the chance of the event (a 3 digit palindrome) / the number of possible 3 digit numbers we get the probability. So, 90/900. 90/900 can be simplified to 1/10. Therefore, A is 1/10.

Now for the final GRE answer, asking if A>B, A<B, A=B, or Cannot be determined (D), we compare the value of A (1/10) to B (1/10) and can see clearly that the answer is C, the two values are equal.

ANSWER: C, the two quantities are equal.

4 digit numbers have nothing to do with this problem. Not sure where you got that from but the question is asking for what the probability is that a 3 digit number would be a palindome. 3 digit numbers start at 100 and end at 999. 099 is a 2 digit numbers and 1000 would be a 4 digit number, so neither is relevant to the question. The probability that a 3 digit number is a palindrome is: how many ways can a 3 digit number be a palindrome /how many 3 digit numbers are there. There is no inconsistency or obfuscatory elements to this problem and therefore D is not going to be a possible answer.

i Think quantity A is the answer

Answer: C
A: The probability chat a 3-digit number chosen at random is a palindrome
A 3-digit number can be between 100 and 999, we will have (999-100+1) = 900 3-digit numbers.
For a 3-digit number to be palindrome no matter what the middle number is, the first and third number should be the same, for instance 101, 111, 121, 131,…,181,191 are all 3-digit palindromes. Here the first and third number were 1, it can be 2,3,4,…,9 so we will have 9*10=90 3-digit palindrome out of all 900 3-digit numbers. Which is 90/900 = 1/10

All 3-digit palindroms:

101, 111, 121, 131,…,181,191
202,212,222,232, ….,292
.
.
909,919, ….., 999

B: 1/10

So A and B are equal.

9/10, 10/10, 10/10=9/10

I got this answer.

Let us find how many 3-digit numbers are palindrome..
so number is _ _ _ => XYX
so X can be any of the 9 digits from 1 to 9- so 9 ways
Y can be any of the 10 digits - so 10 ways
Total = 9*10=90 ways...

Total 3 digit words = 100 to 999 = (999-100)+1=900

Therefore, probability of choosing a 3-digit palindrome = \(\frac{90}{900}=\frac{1}{10}=A=B\).

so both equal

C

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