A group meeting is attended by 16 delegates.
(a)If a 6-person negotiation panel is selected at random from the 16 delegates, what is the probability the negotiation panel consists of 3 delegates from Atlantis and 3 delegates from Zedonia?
Among the 16, how many are from Atlantis and how many are from Zedonia?
11 are from Atlantis and 5 are from Zedonia
Number of ways to choose 3 from each would be:
n=11!/(8!3!) * 5!/(2!3!)
out of total number of ways
N=16!/(6!10!)
Probability: n/N = 75/364
To find the probability of selecting a negotiation panel consisting of 3 delegates from Atlantis and 3 delegates from Zedonia out of a group of 16 delegates, we need to determine the total number of possible panels and the number of panels that meet the given criteria.
First, let's find the total number of possible panels. We need to choose 6 delegates out of the 16, so we calculate the binomial coefficient:
C(16, 6) = 16! / (6! * (16-6)!) = 8008
So, there are 8008 possible panels.
Next, let's find the number of panels that consist of 3 delegates from Atlantis and 3 delegates from Zedonia. Since we need to choose 3 delegates from each group, we calculate the product of the binomial coefficients:
C(8, 3) * C(8, 3) = (8! / (3! * (8-3)!)) * (8! / (3! * (8-3)!)) = 56 * 56 = 3136
Therefore, there are 3136 panels that consist of 3 delegates from Atlantis and 3 delegates from Zedonia.
Now, we can find the probability by dividing the number of panels that meet the criteria by the total number of possible panels:
Probability = Number of panels with the desired criteria / Total number of panels
= 3136 / 8008
≈ 0.3912 or 39.12%
Therefore, the probability that the negotiation panel consists of 3 delegates from Atlantis and 3 delegates from Zedonia is approximately 0.3912 or 39.12%.