Question

A box contains a large number of balls. Some of the balls are red and some are green. The box either contains 40% red balls or 70% red balls. You have to guess the correct percentage. You are allowed to take a random sample of 1 marble to help you make a decision. You draw a red marble. Also, the probability equals 20% that the box contains 70% red balls and 80% that the box contains 40% red balls. What is your best guess? What is the probability you win? Can the problem also be solved using two way table?

Answer 1

To determine the best guess, we need to find the conditional probability that the box contains 70% red balls given that a red marble was drawn. Using Bayes' theorem, we can calculate the probability. The probability of winning is 3.5%. The problem can also be solved using a two-way table.

Step-by-step explanation:

To determine the best guess, we need to find the conditional probability that the box contains 70% red balls given that a red marble was drawn. Let A be the event that the box contains 70% red balls, and B be the event that a red marble was drawn.

Using Bayes' theorem, we can calculate:

P(A|B) = ( P(B|A) * P(A) ) / P(B)

Substituting the given values, we have:

P(70% red|red) = ( 0.2 * 0.7 ) / 0.8 = 0.175

The probability of winning is the probability of correctly guessing the percentage, which is P(70% red|red) * 0.2 = 0.035, or 3.5%.

The problem can also be solved using a two-way table, where the rows represent the possible percentages and the columns represent the outcomes of drawing a red marble or a green marble. The intersection of the row and column represents the probability. In this case, the intersection of the 70% red row and the red marble column is 0.2, which gives the conditional probability of drawing a red marble given that the box contains 70% red balls.