Statistics and probability

Question

1. Assume that there are 12 frozen dinners: 5 pasta, 4 chicken, and 3 seafood dinners. The student selects 5 of them.

What is the probability that at least 2 of the dinners selected are pasta dinners?

2

This problem is involving a weighted dice. For this problem, assume that all the odd numbers are equally likely, all the even numbers are equally likely, the odd numbers are k times as likely as the even numbers, and Pr[4] = 1/18.

What is the value of k?

Thanks in advance

Answer 1

Hey Sean,

(1) It's almost always easier to find "at-least" probabilities by finding their complementary event. For example, the probability of selecting at least one is equal to one minus the probability of selecting none:

P(≥1) = 1 - P(<1) = 1- P(0)

P(≥2) = 1 - P(<2) = 1 - (P(1) + P(0)) = 1- P(1) - P(0)

... and so on ...

To solve your problem, find the probability of selecting zero pasta dinners and the probability of selecting exactly one pasta dinner (which is unfortunately a bit tedious), plug those numbers in, and work out the arithmetic.

(2) Let the random variable X = {1, 2, 3, 4, 5, 6} denote the outcome of rolling one die. Given the weighting, we know:

P(1) = P(3) = P(5) = k

P(2) = P(4) = P(6) = 1⁄18

P(even) = P(2) + P(4) + P(6)

Like question 1, we'll need to use the complementary event to solve for k.

P(odd) = P(1) + P(3) + P(5) = k = 1 - P(even)

Solve by working out the arithmetic.

Hope this helps!