10. Anton, Betty, and Carl have been serving time for grand theft auto (stealing a car)....

Step 1 :
From the information, observe that Andy, Bob and Charley have all been serving time for grand
theft auto.
According to prison scuttlebutt, the warden plans to release two of the three next week.
They all have identical records, so the two to be released will be chosen at random, meaning that
each has a two-thirds probability of being included in the two to be set free.
There are three cases each with 1/3 chance.
Consider A is the event that represents Andy gets released.
Consider B is the event that represents Bob gets released.
Consider C is the event that represents Charley gets released.
Step 2 :
Andy and Bob are set to be released.
The guard must tell Andy that Bob will be released. In this case, 100% of the time, the guard will
tell Andy that Bob was released.
100% of the time. Andy will be the other prisoner released.
Andy and Charley are set to be released.
The guard must tell Andy that charley will be released.
In this case, 100% of the time, the guard will tell Andy that Bob was released.
100% of the time, Andy will be the other prisoner released.
Bob and Charley are set to be released
The guard has a 50% chance of telling Andy that Bob was released, and 50% chance of telling
Andy that Charley was released.
0% of the time, Andy will be released.

Step 3 :
-The probability that the guard says Bob is released, and Andy is also released =1/3.
-That is if the Bob is released first and then Andy has the following probability:P(AB)
Probability of P(A)=1/3

Step 4 :
-The probability that the guard says Bob is released, and Andy is also released = 1/3.
-That is, if Charley is released first, then Andy has the following probability
P(ANC)
Probability =
P(C)
1/3
*1/2
Hence, the concern is not justified.