A die is tossed twice find the probability that;
a. A one and six occurs.
b. At least one three occurs.
c. Both numbers are the same.
Now, the probability of 2nd event is easy to calculate, it is the probability for getting the same number (whatever it might be 1,2,3,4,5 or 6) is 1/6.
The 1st and 3rd event should have same probability because of symmetry. Let that be x.
So, x + 1/6 + x = 1 (because these events are mutually exclusive and exhaustive)
Hence, x = 5/12
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