Question

Palindromes are numbers such as 44 and 383 and 1441 that read the same from right to left as they do from left to right. If a 4-digit number is chosen at random, what is the probability that it will be a palindrome?
(a) 1/100
(b) 1/99
(c) 1/90
(d) 1/10
(e) 1/9

Answer 1

the answer is supposed to be (a) 1/100

Answer 2

it's a

Answer 3

but i don't know how

Answer 4

The first two digits define the palindrome, not counting the first one-digit numbers, we receive \(\frac{1}{90}\)

Answer 5

(If we consider leading zeroes as non-important)

Answer 6

Yes, as @LolWolf said it depends on your definition of 4-digit number, which I believe by convention CANNOT start with zeroes...

Answer 7

In which case it is indeed 1/90

Answer 8

probability= number of the Palindromes / total number that can be made with 4 digit

total number is = 9*10*10*10
number of the Palindromes = 9*10

so probability = 90/9000

Answer 9

Ah yes right! forgot division! My bad.

Answer 10

can u explain how u got the numbers that u multiplied please?

Answer 11

No, wait, you're right, @amir.sat , it is \(\frac{1}{100}\) since the last two digits, which depend on the first two, have a such a probability of equalling the first two.

Answer 12

for total number u gotta multiply the probability of the numbers that van be in each digit

Answer 13

and for the Palindromes u gotta multiply on the probability of the first two digit from the left

Answer 14

If the answer is supposed to be (a), then they are counting 0120 or 0000 as valid . (Sometimes leading zeros are not allowed)

a palindrome will have the form abba where a and b are 0 to 9
once we pick a and b we have picked all 4 digits.
the digits ab start with 00,01,...09,10,11,...,20,...99
there are 100 ab pairs.
on the other hand there are 0000....9999 or 10,000 4 digit numbers
the ratio is 100/10000= 1/100
so 1 out 100 chance of getting a palindrome

Answer 15

ok thanks