Let’s Make a Deal, a classic American ‘60s game show, hosted by Monty Hall, involved
contestants choosing one of three doors, coloured red, blue and green. Behind one of the doors
was a valuable prize, such as a new car. Behind the other two doors was a banana.
After the contestant chooses on the the three dooes Monty Hall (who knows which door has the
prize behind it) always reveals a door (other than the one chosen by the contestant) that has the
banana. Monty now poses the question to the contestant:
‘Do you want to switch dooes or stick with your original choice?’

a) What is the probability of the prize being behind one of the red, blue or green doors. [2 marks]
b) Enumerate all of the combinations of relevant events and draw an event tree or table showing
these event combinations. From this calculate the probability of the contestant winning should
they stick or switch. [8 marks]
c) Derive Bayes Theorem from the Fundamental Rule. [5 marks]
d) Use Bayes Theorem and/or the Fundamental Rule to formally calculate the probability of the
contestant winning given they switch or stick. Be careful to be explicit and exact in your use of
probability notation.

a) Probability of prize in one of the three doors is 1/3
b)
Behind door 1 Behind door 2 Behind door 3 Results if staying Result if switching
1 2 3 #1 Prize
Car Banana Banana Car Banana
Banana Car Banana Banana Car
Banana Banana Car Banana Car

Probability of contestant winning should they stick as seen in the results if staying column is 1/3
Probability of the contestant winning should they switch is given in the results if switching table and is 2/3
c)
The probability of two events A and B happening, P(A∩, is the probability of A, P(A), times the probability of B given that A has occurred, P(B|A).
Therefore: P(A ∩ = P(A)P(B|A)
Equating the two yields: P(B)P(A| = P(A)P(B|A)
Thus: P(A| = P(A) (P(B|A) /P() – Bayes Theorem
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Thanks a lot, I have always been a lover of data, statistics, and metricks