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Jeff will pick a card at random from ten cards numbered 1 through 10. The number on this card will indicate his starting point on the number line shown below. He will then spin the fair spinner shown below (which has three congruent sectors) and follow the instruction indicated by his spin. From this new point he will spin the spinner again and follow the resulting instruction. What is the probability that he ends up at a multiple of 3 on the number line? Express your answer as a common fraction.

lololololololololll  Mar 24, 2018
 #1
avatar+92619 
+2

I just wrote up a table for each of the possible startpoints, there are 10 of them, and then just worked out all the possible paths.

 

the ones that end in a multiple of 3 are

 

123           P(1,2,3) = \(\frac{1}{10}\times\frac{2}{3}\times\frac{2}{3}=\frac{4}{90} \)

343

456

543

656

676

789

876

989

9 10 9

 I have done the first one, work out the probs for all of them and add the probs together for your answer.

 

You need to check what I have done as well :)

Melody  Mar 25, 2018

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