Answer:
Given that;
a)
2 machines, 1 and 2
Machine 1
pays 10% of times when the machine is generous
So the probability that the machine 1 pays given that its generous is
P (Pays/machine 1 is generous) = 10% = 0.10
Machine 2
pays 20% of times when its generous
i.e the probability that the machine pays given that its generous is;
P (Pays / machine 2 is generous) = 20% = 0.20
Also we assume there is equal chance of being generous
i.e
P(machine 1 is generous) = P(machine 2 is generous ) = 0.5
b)
this is to help obtain the probability that machine 1 is generous given that the player loose in the first bet,
i.e P(machine 1 is generous / lost)
c)
Since, the probability that the machine is generous is 0.5,
it can be said that there is 50% chance that machine 1 is generous when the player loses the first bet
d)
therefore The required probability is calculated as;
P(machine 1 is generous/lost) = p(machine 1 is generous ∩ lost) / p (lost)
= [p(lost/machine 1 is generous) × p(machine 1 is generous)] / [{(1-p(pays/machine 1 is generous)) × p(machine 1 is generous)} + {((1-p(pays/machine 2 is generous)) × p(machine 2 is generous)}]
= [(1 - p(pays/machine 1 is hereous)) × p(machine 1 is generous)] × [(( 1- 0.10) × 0.5) + ((1 - 0.20) × 0.5)]
= ( 1- 0.10) × 0.5) / 0.85
= 0.5294
so the probability that the player loses the first bet given that the machine is generous is 0.5294
e)
Since the gotten probability that the player loses the first bet given that machine 1 is generous is close to 0.50 then it can be said that the probability is consistent with the expectations.
