Doubt in the calculation: Consider that the number of customers who buy a T0 apartment is a random variable....

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cmrs

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Consider that the number of customers who buy a T0 apartment is a random variable.

Considering the percentage probability of a customer buying a T0 apartment to be 20%

Suppose 14 customers are randomly selected. What is the probability of at least
least half have bought a T0 apartment knowing that at least 3 customers have
had they done?
What is the probability of finding 3 customers who bought a T0 apartment on two consecutive days?
 
cmrs said:
Consider that the number of customers who buy a T0 apartment is a random variable.

Considering the percentage probability of a customer buying a T0 apartment to be 20%

Suppose 14 customers are randomly selected. What is the probability of at least
least half have bought a T0 apartment knowing that at least 3 customers have
had they done?
What is the probability of finding 3 customers who bought a T0 apartment on two consecutive days?
Click to expand...
You said you have doubts in your calculations. Can you share them?
 
not 0.20 but 0.16
fist part
P(X>=7 | X >=3) <=> (P("X>=7" ∩ "X>=3")) / P(X>=3) <=> P(X>=7)/P(X>=3)
second part
Customers who bought a T0 = 3
(P(X=3) * P(X=0)) + (P(X=2) * P(X=1)) + (P(X=1) * P(X=2)) + (P(X=0) * P(X=3))
Where p=0,16 and n=4
 
Last edited:
cmrs said:
not 0.20 but 0.16
fist part
P(X>=7 | X >=3) <=> (P("X>=7" ∩ "X>=3")) / P(X>=3) <=> P(X>=7)/P(X>=3)
second part
Customers who bought a T0 = 3
(P(X=3) * P(X=0)) + (P(X=2) * P(X=1)) + (P(X=1) * P(X=2)) + (P(X=0) * P(X=3))
Where p=0,16 and n=4
Click to expand...
They look correct assuming each customer and sales on each day are independent. Why do you doubt your calculation?
 
cmrs said:
What is the probability of at least
least half have bought a T0 apartment knowing that at least 3 customers have
had they done?
Click to expand...
Did these at least 3 buy an apartment or not?
 
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