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Somnium
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Help with a question in combinatorics 28 Sep 2018, 01:20
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Hello, would much appreciate help with this kind of question :

there are 3 baskets, each with 5 balls.
in first basket we have 4 black balls, 1 yellow.
in second we have 3 black balls, 2 yellow.
in third we have 2 black , 3 yellow.
we pick one ball from each basket. what is the probability that we get at least one black ball ?

thanks

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Gagoosh
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Help with a question in combinatorics 28 Sep 2018, 01:28
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Somnium
Hello, would much appreciate help with this kind of question :

there are 3 baskets, each with 5 balls.
in first basket we have 4 black balls, 1 yellow.
in second we have 3 black balls, 2 yellow.
in third we have 2 black , 3 yellow.
we pick one ball from each basket. what is the probability that we get at least one black ball ?

thanks
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Is the answer 119/125 ?
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Salsanousi
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Help with a question in combinatorics 28 Sep 2018, 01:43
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Somnium
Hello, would much appreciate help with this kind of question :

there are 3 baskets, each with 5 balls.
in first basket we have 4 black balls, 1 yellow.
in second we have 3 black balls, 2 yellow.
in third we have 2 black , 3 yellow.
we pick one ball from each basket. what is the probability that we get at least one black ball ?

thanks
Show more

Happen to have the answer choices? What is the source for this question?
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Somnium
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Help with a question in combinatorics 28 Sep 2018, 03:33
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in one of official CATs , but unfortunately I don't remember the exact data that was there so I don't know what the answer should be. I changed the stats a bit..
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ccooley
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GMAT 1: 790 Q51 V49
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Help with a question in combinatorics 01 Oct 2018, 12:20
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Somnium
Hello, would much appreciate help with this kind of question :

there are 3 baskets, each with 5 balls.
in first basket we have 4 black balls, 1 yellow.
in second we have 3 black balls, 2 yellow.
in third we have 2 black , 3 yellow.
we pick one ball from each basket. what is the probability that we get at least one black ball ?

thanks
Show more

If you see language like 'at least one' in a probability problem, this strategy will usually work. Instead of finding the probability of getting one black ball, then two black balls, then three black balls, you can actually just find the probability of getting all yellow balls, and subtract from 1. You can do that because you're basically looking for the probability that anything other than all yellow is what happens. That'll be the 'opposite' of the probability that you get all yellow.

The probability of getting all yellow balls is 1/5 * 2/5 * 3/5 (the product of the probability of getting a yellow ball from each basket). That equals 6/125.

So, the probability of getting at least one black (in other words, NOT getting all yellow) is 1-6/125 or 119/125.

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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!