Hi all,
I found out that the solution of problem 197 is not correct.
In the first 3 days has to have 1 to 2 malfunctions to satisfy the third malfunction on the fifth day.
The numerator is : (2 of 4)*0.4^2*0.6^2*0.4 = 0.13824
And the denominator is: 1 - 0.4^3 - 0.6^3 = 0.72 (not having three malfunctions and 0 malfunction)
Then the conditional P = 0.13824/0.72 = 0.192
Any comments and solutions?
Thank you!:oyh:
I found out that the solution of problem 197 is not correct.
In the first 3 days has to have 1 to 2 malfunctions to satisfy the third malfunction on the fifth day.
The numerator is : (2 of 4)*0.4^2*0.6^2*0.4 = 0.13824
And the denominator is: 1 - 0.4^3 - 0.6^3 = 0.72 (not having three malfunctions and 0 malfunction)
Then the conditional P = 0.13824/0.72 = 0.192
Any comments and solutions?
Thank you!:oyh:
Problem 197 in exam P sample questions
0 commentaires:
Enregistrer un commentaire