Vera has a pot containing 4 red grapes and 6 green grapes. She takes a grape at random and eats is. She then takes another grape at random and easts it. What probability should be written in place of Z in the probability tree __

Question

Question

Vera has a pot containing 4 red grapes and 6 green grapes. She takes a grape at random and eats is. She then takes another grape at random and easts it. What probability should be written in place of Z in the probability tree __

Answer 1

The probability for 'Z' placeholder from the probability tree should be \(\frac{1}{2}\).

Answer 2

## Step 1: To solve this problem, I will construct a probability tree and figure out the unknown probability identified as `Z`. ## Step 2: There are two paths that realize taking two grapes, one is 'red then green', and the other is 'green then green'. For the first event, both outcomes are directly stated: the probability of picking a red grape first is \(\frac{4}{10} = \frac{2}{5}\), and the probability of picking a green grape first is \(\frac{6}{10} = \frac{3}{5}\). ## Step 3: For the second event, if the first grape is red, there are 9 grapes left; 6 of them are green and 3 of them are red. Therefore, the probability of picking a green grape second if the first grape eaten was red should be \(\frac{6}{9} = \frac{2}{3}\). Thus the first branch (red, green) is unchanged from the available choices given. ## Step 4: Let's assume there is only one option for the second part of the event (after taking the first green grape) labeled as a `Z`. Because we assumed the first grape Edgar took was green then there are now 9 left in the pot. Hence, the only conceivable `Z` are for picking the second grape which can be Red or Green. But if those were what was meant to display originally, then green grapes would succeed their appearance which made the `Z` label actually refers to the second green branch. ## Step 5. As a consequence, `Z` must signify the chance of choosing a green grape if the first choice was green. So, Logic dictates now only 8 grapes left. 4 of them are green- meaning the likelihood second green pick should be \(\frac{4}{8} = \frac{1}{2}\).

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Vera has a pot containing 4 red grapes and 6 green grapes. She takes a grape at random and eats is. She then takes another grape at random and easts it. What probability should be written in place of Z in the probability tree __

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Question

Vera has a pot containing 4 red grapes and 6 green grapes. She takes a grape at random and eats is. She then takes another grape at random and easts it. What probability should be written in place of Z in the probability tree __ __ __

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Explain

Hot Questions

Vera has a pot containing 4 red grapes and 6 green grapes. She takes a grape at random and eats is. She then takes another grape at random and easts it. What probability should be written in place of Z in the probability tree __