1 Answer

Judie · Answer 1 · 2023-10-11T02:58:38+0000

First let's create the sample space of the experiment

let S denotes the sport car , T denote the pick up truck and R denotes the grandma ride.

Then the possible combinations will be

$\begin{gathered} (S,New),(S,Good),(S,Poor),(S,Fair), \\ (T,New),(T,Good),(T,Poor),(T,Fair), \\ (R,New),(R,Good),(R,Poor),(R,Fair), \end{gathered}$

Now let E deontes the event of getting grandma ride and F denotes the event of getting new vehicle

Then

$n(E\cap F)=1$

So

$P(E\cap F)=(1)/(12)$

So the probabilty of getting a grandma ride and a vehicle in new condition is 1/12

Now let A denote the event of getting a sports car and B denote the event of getting a vehicle in fair condition so

$n(A\cup B)=6$

So

$P(A\cup B)=(6)/(12)$

So the probabilty of getting a sports car or a vehicle in fair condition is 6/12

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