# Probability distributions help!?

Suppose that a die is weighted such that prime numbers are twice as likely to appear as non-primes. What is the probability of rolling 5 at most 2 times in 15 tries? (recall: 1 is neither prime nor composite!) What is the expected value?

### 1 Answer

- PuzzlingLv 73 years agoFavorite Answer
The final probability would be equivalent to having a 9-sided die with 2, 3 and 5 each appearing twice compared to 1, 4 and 6 appearing once.

P(1) = 1/9

P(2) = 2/9

P(3) = 2/9

P(4) = 1/9

P(5) = 2/9

P(6) = 1/9

Next, use the Binomial Probability formula:

P(X = k) = C(n, k) * p^k * q^(n-k)

n : number of trials 15

k : number of successes (0, 1 or 2)

p : probability of success (2/9)

q : probability of failure (1-p = 7/9)

So compute P(X = 0) + P(X = 1) + P(X = 2):

P(X = 0) = C(15,0) * (2/9)^0 * (7/9)^15

P(X = 1) = C(15,1) * (2/9)^1 * (7/9)^14

P(X = 2) = C(15,2) * (2/9)^2 * (7/9)^13

P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)

= C(15,0) * (2/9)^0 * (7/9)^15 + C(15,1) * (2/9)^1 * (7/9)^14 + C(15,2) * (2/9)^2 * (7/9)^13

= 65787638066353 / 205891132094649

≈ 31.95%

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