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# A reality show has four answer choices for each question - A, B, C, an

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Q51  V47
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Re: A reality show has four answer choices for each question - A, B, C, an [#permalink]
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Abhibarua wrote:
A show has four answer choices for each question - A, B, C, and D. The participant, John, correctly knows that the right answer is either A or D. The show has a '50:50' option by using which participants can eliminate two wrong answer choices. What is the probability that John uses the '50:50' option and is still left with both A and D?

A. 1/4
B. 1/5
C. 1/6
D. 1/10
E. 1/ 12

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]
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Re: A reality show has four answer choices for each question - A, B, C, an [#permalink]
IanStewart wrote:
Abhibarua wrote:
A reality show has four answer choices for each question - A, B, C, and D. The participant, John, correctly knows that the right answer is between A and D. The show has a '50:50' option by using which participants can eliminate two wrong answer choices. What is the probability that John uses the '50:50' option and is still left with answer choices A and D?

A. 1/4
B. 1/5
C. 1/6
D. 1/10
E. 1/ 12

This is a good question concept, but there are several problems with the question as written. Most importantly, the right answer isn't among the choices. When the game show eliminates two answers, it must eliminate two of the three wrong answers. So it must leave one of the three wrong answers 'on the board'. So there's a 1/3 chance they'll leave up the wrong answer John is worried about, and a 2/3 chance they'll leave up one of the other two wrong answers, and the answer is 1/3. If that's a bit conceptual, you can imagine that the right answer is actually A. Then the game show is obligated to leave exactly one of B, C or D on the board, so there's a 1/3 chance they leave D up. The same is true when the right answer is D; then there is a 1/3 chance they leave A up. So no matter what answer is right, there's a 1/3 chance A and D are still available after the 50-50 is used.

But I think the question means to describe a "game show" (not a "reality show") and they really shouldn't say "between A and D" after listing the options in order. If we say "x is an integer between 1 and 4", we mean "x is either 2 or 3". We do not mean "x is one of the two numbers 1 or 4", but that is (I assume) the intended meaning in this question. What is the source?

Thanks for your feedback. I edited the question for better clarity.
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Q51  V47
Re: A reality show has four answer choices for each question - A, B, C, an [#permalink]
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).
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Re: A reality show has four answer choices for each question - A, B, C, an [#permalink]
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.
Re: A reality show has four answer choices for each question - A, B, C, an [#permalink]
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