Welcome to Westonci.ca, where curiosity meets expertise. Ask any question and receive fast, accurate answers from our knowledgeable community. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.
Sagot :
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.
Thanks for using our service. We aim to provide the most accurate answers for all your queries. Visit us again for more insights. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.
