Posted by **Arun** on Thursday, March 1, 2012 at 12:32am.

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**, Thursday, March 1, 2012 at 9:02am
"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 not-prime 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

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