IanStewart wrote:
Abhibarua wrote:
There are 6 ways to pick 2 from 4 answer choices [4!/(2!*2!)].
These six combinations are AB, AC, AD, BC, BD, CD. However, BC is not an allowed combination because it does not contain the right answer, which is either A or D.
So, the number of desired outcomes (AD) = 1
the number of possible outcomes = 5
So, the probability that John uses the '50:50' option and is still left with both A and D = 1/5 [Option: B]
That solution is not correct. It's true that BC cannot be the remaining two answers. But when the correct answer is A, it is also true that BD and CD cannot be the remaining two answers. And when the correct answer is D, then AB and AC cannot be the two remaining answers. You haven't accounted for that anywhere.
The game show is obligated to leave the right answer available after the 50-50 is used. The game show is always choosing 2 wrong answers from the 3 wrong answers to the question. So from the game show's point of view, when it randomly eliminates two wrong answers, it doesn't have six choices. It only has three choices (3C2 = 3).
You are right. After the '50:50', the answer choices can only have 3 possible combinations. (1 way to choose the right answer * 3 ways to choose 1 from 3 wrong answers) [1 * (3!/1!*2!)]
For example, if the right answer is A, the possible combinations are AB, AC, and AD.
If the right answer is D, the possible combinations are AD, BD, CD.
Either way, the number of desired outcomes (AD) is 1 and the number of possible outcomes is 3. So the probability should be 1/3.
Thanks again.