Question

5. Consider the Monty Hall Problem. A game show host shows you three doors, and indicates that one of them has a car behind it, while the other two have goats. You win a car if you end up choosing a door with a car behind it. The game is conducted as follows: • You pick an initial door out of the three available. • The game show hosts then opens a door (out of the remaining two doors) with a goat.

Answer 1

Solo given that Considering the the data is mority Hall Pi A game show host shows you three doors, and indicates that one of them has a car behind it, while the other two have goats det Do be the the event { "Door : contains a carmy and Ribe { " The game hosts rencals a goat behind dool So, let us consider gate 1 initially and morty shows up us gat a, Now I'll find the probability up winning if we dont switch there we will be using Baye's theorem to find the Conditional Probabilities PCR₂10,) & PCO) P/ PIR2) = P(R1 DD x P (D1) + P(R21 D₂) x P (D₂) + p (R2 103 ) x P (D₃ 12 x 1/3 toxt + 1 + 1 / 2 | 1/6 - 3 6 +0+ 5 situation in this Hence, its better to make a switch now we'll deal already gate 1 Here with we've chosen two cases

(1) it morty opens gate e and as ir monty ogens gate 3 - axe will And the probability of success when we don't make a switch in both cases Care (1) ip monti. y opens the gale a PCR, 10,) K P (D) Pl Dildo PC Rolo, ) * P (o) + P(R 102 x P (D₂) + P (R 10₃) + P (Dz. 0.5x0.5 to toittlxool 0.2 oro = 0.714285 (ii) & monity opens opens go gate 3: P(Rolpox Pro) PORD,) xP(0,)+P (R310) xp(02)+(R3102) 7P() B (P. 12) = 0.5X0.5 0.540.5tox0.4+1xool 0.25 + 0 + 0. 0.714285 Thus we should not . Case ic for behid door 2.40% of of the tome Switch behid door the time the gates in this 150% of the time, & behind door 3 10 %