Posted by Arun on .
The probability that at most 3 tosses of a balanced dice are required to get a prime number on top is equal to
a. 7/8, b.1/4, c. 1/2, d.3/4.

Probability and statistics 
MathMate,
"at most 3" = three or less
So the prime number can appear at least once in the three tosses, no matter which one(s).
Prime numbers ≤6 are 2,3 and 5.
So the outcome of each toss is either prime or not prime.
The probability of getting a prime is 3 out of 6 possible events, or 3/6=1/2.
We look for at least one success out of three tosses. Thus the only situation failure can occur is all three tosses are notprime numbers, which has a probability of P(0)=(1/2)^3.
So the probability of getting at least one success is the complement, or P(1,2,3)=1(1/2)^3 = 7/8