Of 25 calculators, 4 have faulty displays. A random sample of 2 calculators is selected
without replacement.
Let C1 be the event in the first calculator selected the display works
Let N2 be the event in the second calculator selected the display fails
What is Pr(first display works and second display fails)?
Answer
= Pr(C1∩N2)
= Pr(C1).Pr(N2\C1) = (21/25).(4/24) = 0.14. (using the multiplication rule for the probability
of two events)
*actually i wanna ask about Pr(N2\C1)=(4/24) this value 4/24 i cannot understand it someone else can explain to me how the value come from.and show me the easy step can get 4/24 i really want to understand it about this ~~thank all><
grateful~someone can help me thank a lot
without replacement.
Let C1 be the event in the first calculator selected the display works
Let N2 be the event in the second calculator selected the display fails
What is Pr(first display works and second display fails)?
Answer
= Pr(C1∩N2)
= Pr(C1).Pr(N2\C1) = (21/25).(4/24) = 0.14. (using the multiplication rule for the probability
of two events)
*actually i wanna ask about Pr(N2\C1)=(4/24) this value 4/24 i cannot understand it someone else can explain to me how the value come from.and show me the easy step can get 4/24 i really want to understand it about this ~~thank all><
grateful~someone can help me thank a lot