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I find a bowl of colored Christmas candies​ (only red and​ green) that were hidden from me in the house one day. I quickly grab a handful before being caught. Before shoving them into my mouth to get rid of the​ evidence, I notice that I have selected 5 red candies and no green ones.​ Let's say there were 6 green and 7 red candies in the bowl before my discovery. What is the probability of this event​ occurring?

User Nezroy
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1 Answer

3 votes

Answer:

The probability of the occurring the event is 0.01632 approximately

Explanation:

Consider the provided information.

We have selected 5 red candies and no green ones.​ Let's say there were 6 green and 7 red candies in the bowl before discovery.

Therefore the total number of candies are 6+7=13

I have selected 5 red candies out of 7 red candies.

Thus, the probability of selecting 5 red candies and no green candies is:


(^7C_5* ^6C_0)/(^(13)C_5)


(\left((7!)/(5!2!)\cdot(6!)/(6!)\right))/((13!)/(5!8!))\approx 0.01632

Hence, the probability of the occurring the event is 0.01632 approximately