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Math Challenge II-A Combinatorics
Lecture 8, Problem 8.5 c)
There is no fixed answer to this problem, because the primes are 2, 3, 5, and 7. You are only given that the probability of getting 1, 3, or 5 is 0.15, the probability of getting 2 or 4 is 0.05, the probability of getting 6 is 0.25, and the probability of getting 7 is the same as the probability of getting 8. Besides, the solution in the book only says that the primes are 2, 3, 5, 7 and doesn't say anything else.
Since $\Omega=\{1,2,3,4,5,6,7,8\}$, we have that $$P(1) + P(2) + P(3) + P(4) + P(4) + P(5) + P(6) + P(7) + P(8) = 1.$$ We are given that $P(1) = P(3) = P(5) = 0.15$, $P(2) = P(4) = 0.05$, $P(6) = 0.25$ and $P(7) = P(8)$.
From these, in part (a) we find that $P(7) = P(8) = 0.1$. Now we know the probabilities of picking each individual number.
Then, for part (b), we can find the probability of picking an even number by adding the probabilities of getting each of the even numbers, that is, $P(2) + P(4) + P(6) + P(8) = 0.45$.
Finally, for part (c), to find the probability of picking a prime number we proceed as in part (b), this time we need to add the probabilities of picking each prime number, which are $2$, $3$, $5$, and $7$, that is, $P(2) + P(3) + P(5) + P(7) = 0.45$.
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