Question

NO LINKS!!! THIS IS NOT MULTIPLE CHOICE!!!

Answer 1

Answer:

Part (a)

see attached for sample space

Part (b)

$\sf Probability\:of\:an\:event\:occurring = \dfrac{Number\:of\:ways\:it\:can\:occur}{Total\:number\:of\:possible\:outcomes}$

Total number of possible outcomes = 12

$\textsf{P(Renae takes an odd-numbered bus)}= \dfrac{9}{12}=\dfrac34$

$\textsf{P(Renae does not write a letter)}= \dfrac{8}{12}=\dfrac23$

$\textsf{P(Renae catches no.28 bus then reads a book)}= \dfrac{1}{12}$

Part (c)

Her activity does not depend on which bus she takes, as we have been told that once she is on the bus, she randomly selects the activity.  Therefore, the activity selection is independent from the bus selection.

The sample space confirms this as it shows that the probability of the selecting each activity is equally likely for each of the 4 buses.