Bunuel wrote:

minwoswoh wrote:

Hi Bunuel,

One more comment...

I think that even if the question were to tell us that are indeed more than 1 green chip, the answer will still be E.

Example 1: 8 red chips and 2 green chips

Probability of both green = 2/8 * 1/7 = 1/28

Example 2: 12 red chips and 3 green chips

Probability of both green = 3/12 * 2/11 = 1/22

Example 3: 16 red chips and 4 green chips

Probability of both green = 4/16 * 3/15 = 1/20

Can we then conclude that the more quantity overall, the more chances you have of getting a green chip?

It should be:

Example 1: 8 red chips and 2 green chipsProbability of both green = 2/10 * 1/9 = 1/45

Example 2: 12 red chips and 3 green chipsProbability of both green = 3/15 * 2/14 = 1/35

Example 3: 16 red chips and 4 green chipsProbability of both green = 4/20* 3/19 = 3/95

The probability of choosing 1 green chip if the ratio of red to green is 4:1 will be the same (1/5) no matter how many total chips we have.

The probability of choosing 2 green chips will increase by increasing the total number of chips.

Hope it's clear.

Such a careless mistake from my part!

It´s crystal clear now, Bunuel. This was very helpful for me to understand how DS tests the concept of Ratios vs. Concrete Numbers in Probability (since obviously, probability is expressed as a ratio itself...)

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