On a game show, there are 10 keys in a bag, and 3 of the keys start a car. a contestant randomly chooses a key. it does not start the car. she returns the key to the bag, the host mixes up the keys, and she randomly selects another key. this key does not start the car either. what is the probability of this no start, no start outcome? write the answer as a percent. %

First of all, as I told you a little while ago, here is the formula
that solves ALL probability problems.  Memorize this:

Probability =
       (number of different successful possibilities)
divided by
       (total number of all possibilities).

There are 10 keys in the bag.  The total number of possible results
of pulling out one key out is 10.

3 of the keys start the car.  The other 7 keys don't.

What are the 'successful' possibilities ? 
The question asks for the probability of pulling keys that DON'T work. 
So for this particular calculation, we're talking about choosing keys
that DON'T work.  There are 7 of those.

The probability of drawing a key that doesn't work the first time is 7/10 .

Since the first key is put back in the bag and the bag is mixed, the
situation on the second draw is exactly the same as on the first one.

Thje probability of drawing a key that doesn't work the second time is 7/10 .

The probability of both events happening is (7/10)x(7/10) = 49/100 = 49% .

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