Question

A locker combination has three nonzero digits, with no digit repeated. If the first two digits are odd, what is the probability that the third digit is also odd?

Answer 1

Given the first two digits of a 3-digit combination are odd (5 options each), only 3/5 of the remaining combinations will also have an odd third digit (5 total options).

There are three places for the digits in the combination. Since the first two must be odd, and non-repeating, there are 5 choices for the first digit (1, 3, 5, 7, or 9) and then 4 choices for the second digit (the remaining odd digits).

The third digit can be any of the five remaining digits (odd and even) because we are only concerned with the first two being odd.

So, there are a total of 5 * 4 * 5 = 100 combinations of which the first two digits are odd. However, only 5 * 4 * 3 = 60 of these will have all three digits be odd.

Therefore, the probability that the third digit is odd given that the first two are odd is 60/100 = 3/5.

Answer 2

If you can get this using evens,
The digits are probablly 1 to 9, then the even is 2,4,6,8 and the odd are 1,3,5,7,9.

Prodd(ways/choices)=5/(2+5)

So, Using odd numbers, The digits are likely to be 1-9, so the odds are 1,3,5,7,9. Look at the middle number, 5⇒6+3 (because of odds not evens that use 2). 6,7,8,9. 5/9 is your answer.


Maybe this helped and if it did then yw. :D