Question

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?

Answer 1

Answer:

The probability of the occurring the event is 0.01632 approximately

Step-by-step 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:

$\frac{^7C_5\times ^6C_0}{^{13}C_5}$

$\frac{\left(\frac{7!}{5!2!}\cdot\frac{6!}{6!}\right)}{\frac{13!}{5!8!}}\approx 0.01632$

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