***Please answer both A and B Question 11: On a TV game show, a contestant is se
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***Please answer both A and B
Question 11: On a TV game show, a contestant is selected randomly from the audience, and offered a chance to choose 1 of 3 closed doors. Behind 1 of the doors is a fantastic prize, while there is nothing behind the other 2 doors. Once the contestant makes her choice, the host (who knows where the prize is) always opens one of the other 2 doors, intentionally revealing an empty door. This leaves 2 remaining doors - the contestant's choice, and one other - with the prize behind 1 of those doors. Before revealing the outcome, the host offers the contestant one last choice: she can stick with his original choice, or else a) switch to the other remaining door. In order to have the greatest probability of winning the prize, what should the contestant do? Justify your answer by calculating the probabilities of winning, for eaclh option (Hint: there are 3 possible answers to choose from: either it is better to stick, or it is better to switch, or it makes no difference) If the host didn't know where the prize was, then opening one of the non-selected doors would pose a problem. On average, how often would this problem arise? b)Explanation / Answer
Ans a To have a greater probability to win the prize the contestant should switch the doors and the reason is explained below :-
Suppose that we take one case that the contestant chooses Gate 1 and the host opens Gate 3 because he knew that there is no prize behind that door and any other case can be explained similarly.
Lets calculate the probability now (that is after Gate 3 has been opened that the prize is behind Gate 1 or Gate 2
Behind Gate 1 - Now P(Gate 3 opened| Prize is behind Gate 1 ) is 0.5 because host could have opened Gate 2 or 3 since prize is behind Gate 1. P(Prize behind Gate 1 | Gate 3 opened and Gate 1 chosen) = P(Prize behind Gate 1 and Gate 3 opened and Gate 1 chosen)/P(Gate 3 opened and Gate 1 chosen) = P(Gate 3 opened | Prize behind Gate 1 and Gate 1 chosen)* P(Prize behind Gate 1 and Gate 1 chosen)/P(Gate 3 opened | Gate 1 chosen) * P(Gate 1 chosen) = (0.5 * 1 / 3) / 0.5 = 1/3
P(Prize behind Gate 1 and Gate 1 choosen) is 1/3 because the contestant doesn't know where the prize so both are independent events and P(Gate 1 chosen) = 1 and P(Prize behind Gate 1) = 1/3 and P(Gate 3 opened | Gate 1 chosen) = 0.5
Similarly We can calculate that Behind Gate 2 probability is 2/3 .Here P(Gate 3 opened| Prize is behind Gate 2 ) = 1 because the host has to choose Gate 3 he has no option.
Thus it is better to switch because the probability of prize behind the other gate is higher.
Ans b) Suppose If Gate 1 was chosen then
If prize is behind gate 1 then there would be no problem arising to choose a gate without prize but if prize is behind any other gate then the probability of choosing a gate without prize is 0.5 .
Thus on an average the probability of choosing a gate without prize = 1/3 * ( 1 + 0.5 + 0.5) = 2/3 where 1/3 is the probability of choosing a gate followed by the probability according to two cases mentioned above.
Thus on a average one - third time the trouble could be caused.
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