How do they compute probability here?
Dear Student,
Solution)
When a die is thrown, then probability of getting a six =
then, probability of not getting a six =
If the man gets a six in the first throw, then
probability of getting a six =
If he does not get a six in first throw, but gets a six in second throw, then
probability of getting a six in the second throw =
If he does not get a six in the first two throws, but gets in the third throw, then
probability of getting a six in the third throw =
probability that he does not get a six in any of the three throws =
In the first throw he gets a six, then he will receive Re 1.
If he gets a six in the second throw, then he will receive Re (1 - 1) = 0
If he gets a six in the third throw, then he will receive Rs(-1 - 1 + 1) = Rs (-1),
that means he will lose Re 1 in this case.
Expected value =
So, he will loose Rs.
Regards!