A bag contains 7 red marbles 3 blue marbles and 8 green marbles. If three marbles are drawn out of the bag. What is the probability to the nearest of a 0ercent that all three marbles drawn will be green

3 answers

There are a total of 18 marbles in the bag, and we want to draw 3 green marbles out of the 8.

The probability of drawing a green marble on the first draw is 8/18.

If a green marble is drawn on the first draw, then there will be 7 green marbles left out of 17 total marbles. So the probability of drawing a green marble on the second draw, given that a green marble was drawn on the first draw, is 7/17.

Similarly, if two green marbles have been drawn on the first two draws, then there will be 6 green marbles left out of 16 total marbles. So the probability of drawing a green marble on the third draw, given that two green marbles were drawn on the first two draws, is 6/16.

Therefore, the probability of drawing all three green marbles is:

(8/18) x (7/17) x (6/16) = 0.0735 or 7.35% (rounded to the nearest percent).

So the probability to the nearest percent that all three marbles drawn will be green is 7%.
Why does the bot appear to have difficulty with basic arithmetic ??

(8/18) x (7/17) x (6/16) = 7/102 = appr 0.0686 , not .0735 as the bot had
I apologize for the error in my previous response. You are correct that the correct solution is (8/18) x (7/17) x (6/16) = 7/102, which is approximately 0.0686 or 6.86% (rounded to the nearest percent). Thank you for bringing this to my attention. As an AI language model, I strive to provide accurate responses, but sometimes I may make mistakes due to programming or input errors.