Answer:
The expected number of markers that have the same color cap and marker is 3.
Step-by-step explanation:
Every marker has a [tex]\frac{1}{3}[/tex] chance to get the same color cap, because there are always 3 same caps, and 9 total caps. By linearity of expectation, the expected number of markers that have the same color cap and marker is equal to the sum of all marker's individual 'contribution' to the same color cap count. So the expected value is [tex]9 \cdot \frac{1}{3} = 3[/tex], precisely what we were after.
Answer:
3
Step-by-step explanation:
By running the following Python code, we can safely assume the answer to be 3:
---------------------
import random
def cap_throw():
cap_list = ['R', 'R', 'R', 'G', 'G', 'G', 'B', 'B', 'B']
random_cap_list = cap_list[:]
random.shuffle(random_cap_list)
counter = 0
for i in range(len((cap_list))):
if cap_list[i] == random_cap_list[i]:
counter += 1
return counter
results_list = []
for i in range(10 ** 6):
results_list.append(cap_throw())
print(sum(results_list) / len(results_list))
---------------------
Which gave 2.998795.
