Asked by Arun
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.
a. 7/8, b.1/4, c. 1/2, d.3/4.
Answers
Answered by
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 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
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
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.