Answer:
The probability that the battery in a critical tool fails at 32 hours was from manufacturer 2 is 0.625
Step-by-step explanation:
Given that:
60% of the batteries are from manufacturer 1
90% of these batteries last for over 40 hours
Let the number of the battery duration be n = 0.90
Therefore n' = 1 - 0.90 = 0.10
Let p = manufacturer 1 and q = manufacturer 2
q = 1 - p
q = 1 = 0.6
q = 0.4
Thus ; 40% of the batteries are from manufacturer 2
However;
Only 75% of the batteries from manufacturer 2 last for over 40 hours.
Let number of battery duration be m = 0.75
Therefore ; m' = 1 - 0.75 = 0.25
A battery in a critical tool fails at 32 hours.
Thus; the that the battery in a critical tool fails at 32 hours was from manufacturer 2 is:
[tex]= \dfrac{q \times m' }{ p \times n' + q \times m' }[/tex]
[tex]= \dfrac{0.4 \times0.25 }{ (0.6 \times 0.1) + (0.4 \times 0.25 ) }[/tex]
[tex]=\dfrac{0.1}{0.06+ 0.1}[/tex]
[tex]=\dfrac{0.1}{0.16}[/tex]
= 0.625
The probability that the battery in a critical tool fails at 32 hours was from manufacturer 2 is 0.625
