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  1. Prepare
  2. Tutorials
  3. 10 Days of Statistics
  4. Day 4: Geometric Distribution II
  5. Discussions

Day 4: Geometric Distribution II

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109 Discussions

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  • illogicalsleepi1
     + 0 comments
    num,den= map(int, input().split())
k=int(input())
p=num/den
geodis=0
for k in range(1,6):
    geodis+=((1-p)**(k-1))*p
print(round(geodis,3))

  • n4m774r_d
     + 0 comments
    n,d = map(float,input().split());k = int(input());p = n/d
print(round(sum((1 - p)**i * p for i in range(k)),3))

  • glamacha
     + 0 comments

    I found the tutorial brining in negative binomials a bit confusing. Isn't a clean way to compute this simply that the probability of not producing a defect in n=5 rounds covers the only possibility other than a defect occurring during the first n=5 inspections?

    python n = 5; p = 1 / 3.0; q = 1.0 - p; print(f"{(1.0-q**n):.3f}")

    // or in C double p = 1.0/3.0; double q = 1 - p; printf("%.3f", 1 - pow(q, 5));`

  • mateuszampili
     + 0 comments
    #The formula for the geometric distribution probability: (P(X = k) = (1-p)^{k-1}p),
# Define the probability of success on a single trial
p = 1/3

# Initialize the probability of finding the first defect in the first 5 inspections
prob = 0

# Loop over the first 5 inspections
for k in range(1, 6):
  # Calculate the probability of finding the first defect on the k-th inspection
  prob_k = (1-p)**(k-1) * p

  # Add this probability to the total probability
  prob += prob_k

# Print the final probability
print(round(prob, 3))

  • oksana_vityuk
     + 0 comments
    p_defect_product = 1/3

n = 5
p_first_defect = 0
for i in range(1, n+1):
    p_first_defect += pow((1 - p_defect_product), i-1)*p_defect_product

print(round(p_first_defect, 3))