Plz plz plz help help❤️ With atleast some
1 answer:
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Answer:
0.999987
Step-by-step explanation:
Given that
The user is a legitimate one = E₁
The user is a fraudulent one = E₂
The same user originates calls from two metropolitan areas = A
Use Bay's Theorem to solve the problem
P(E₁) = 0.0131% = 0.000131
P(E₂) = 1 - P(E₁) = 0.999869
P(A/E₁) = 3% = 0.03
P(A/E₂) = 30% = 0.3
Given a randomly chosen user originates calls from two or more metropolitan, The probability that the user is fraudulent user is :
= 0.999986898 ≈ 0.999987
Answer:
0.8973
Step-by-step explanation:
Relevant data provided in the question as per the question below:
Free throw shooting percentage = 0.906
Free throws = 6
At least = 5
Based on the above information, the probability is
Let us assume the X signifies the number of free throws
So, Then X ≈ Bin (n = 6, p = 0.906)
Now
The Required probability = P(X ≥ 5) = P(X = 5) + P(X = 6)
= 0.8973