A jar contains 6 red jelly beans, 6 green jelly beans, and 6 blue jelly beans.
If we choose a jelly bean, then another jelly bean without putting the first one back in the jar, what is the probability that the first jelly bean will be green and the second will be green as well?

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Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ Calculate the probability of picking the first green jelly bean. ‖ ‖ step 2 ⋮ There are 6 green jelly beans out of 18 total, so the probability is $ \frac{6}{18} $. ‖ ‖ step 3 ⋮ Calculate the probability of picking the second green jelly bean without replacement. ‖ ‖ step 4 ⋮ After one green jelly bean is taken, there are 5 green jelly beans left out of 17 total, so the probability is $ \frac{5}{17} $. ‖ ‖ step 5 ⋮ Multiply the probabilities from step 2 and step 4 to find the combined probability. ‖ ‖ step 6 ⋮ The combined probability is $ \frac{6}{18} imes \frac{5}{17} = \frac{1}{3} imes \frac{5}{17} = \frac{5}{51} $. ‖ ‖ step 7 ⋮ Simplify the fraction $ \frac{5}{51} $ to get the final probability. ‖ ‖ step 8 ⋮ The simplified fraction is $ \frac{5}{51} = \frac{1}{10.2} $, which is approximately 0.09804 or 9.804%. ‖ ∻Answer∻ ⚹ The probability that the first jelly bean will be green and the second will be green as well is approximately 9.804%. ⚹ ∻Key Concept∻ ⚹Probability of consecutive events without replacement⚹ ∻Explanation∻ ⚹To find the probability of two consecutive events without replacement, multiply the probability of the first event by the probability of the second event, considering the change in total number of outcomes after the first event.⚹◊What is the formula for calculating the probability of picking two green jelly beans consecutively without replacement from the jar?⍭ Generate me a similar question◊

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