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 4 on the number line? Express your answer as a common fraction.
The number of possible paths is 10*3*3 = 90
You need to end on 0, 4, 8 or 12 no other multiples of 4 are possible
Say you choose the number 1 the only multiple of 4 you can get to is 0
1 -1 +0 or 1+0-1 that is 2 favourable outcomes if you start with the number 1
Now do the same with all the other starting numbers to work out the number of favourable paths.
then the prob will be number of favourable outcomes / 90
Please let guest finish this on their own