Question

Garth has a chest which contains 5 silver balls and 6 blue balls. He draws balls one at a time without replacement untl he draws a silver at which point he stops. Determine the probability that he draws at least 1 of the blue balls before stopping:

Answer 1

The probability of Garth drawing at least one blue ball before drawing a silver ball is ⅖, using the concept of complementary probability.

Step-by-step explanation:

The question involves calculating the probability that Garth draws at least one blue ball before drawing a silver ball from a chest of balls. To handle this probability question, we use the concept of complementary probability because it is easier to find the probability of the opposite event—that Garth draws a silver ball first— and then subtract it from 1 to get our desired probability.

Firstly, there are 5 silver balls and 6 blue balls, making a total of 11 balls in the chest. The probability that Garth draws a silver ball first is ⅕ (5 silver balls out of 11 total balls). Therefore, the probability that he draws at least one blue ball before a silver ball is 1 - ⅕ which simplifies to ⅖. So, Garth has a ⅖ probability of drawing at least one blue ball before stopping.