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# Jeff will pick a card at random from ten cards numbered 1 through 10.

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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.

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Sep 8, 2018

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Jeeze where are you getting these problems?

First off the two spins have the following distribution on the resulting movement.

Movement to the right is positive, movement to the left is negative.

$$P[(-2,~0,~2)] = \left(\dfrac 1 9,~\dfrac 4 9,~\dfrac 4 9\right)$$

Second we note the possible multiples of 3 which are (0, 3, 6, 9, 12)

So with probability 1/9 of ending up on a multiple of 3 we can choose (2,5,8)

With probability 4/9 of ending up on a multiple of 3 we can choose (3, 6, 9)

With probability 4/9 of ending up on a multiple of 3 we can choose (1, 4, 7, 10)

All cards have probability 1/10 of being chosen and we end up with the following probabilities

$$P[\text{ending up on multiple of 3}] = \dfrac{3}{10} \dfrac 1 9 + \dfrac{3}{10}\dfrac 4 9 + \dfrac{4}{10}\dfrac 4 9 = \dfrac{31}{90}$$

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Sep 8, 2018