Question

A spinner and 6 cards are shown below: A spinner with 3 equal sectors is shown in the figure. The colors Green, Blue, and Red are marked on it. The arrow points at the green color. Six cards are shown on the right side. The colors Purple, Green, Red, Yellow, Blue, and Pink are marked on them. Jordan spins the spinner and picks up a card without looking. What is the probability that the spinner stops at green and a yellow card is selected? 1 over 36 1 over 18 1 over 6 1 over 3

Answer 1

This is a case of joint probability: you want to know what the odds are of this event AND that event occurring. For a joint probability, the formula is usually given as:

P(A and B)=P(A)*P(B)

where P(A) is the probability of event A happening, P(B) is the probability of event B happening, and P(A and B) is the probability of both happening.

For this problem, let's say that event A is the probability that the spinner stops at the green color. Since the spinner has 3 colors and they have equal area (sector) occupied in the circle, each color has the same chance, 1/3 chance of being picked. Thus, there's a 1/3 chance of the spinner stopping at the green color so P(A)=1/3.

Event B is the probability of Jordan picking up the yellow card. There are six possible colors to pick and since Jordan picked up the card without peeking, each color has an equal chance to be picked. Thus, there's a 1/6 chance that the yellow card will be selected so P(B)=1/6.

Using the formula, we can then calculate P(A and B), the probability that the spinner stops at green and a yellow card is selected:

P(A and B)=P(A)*P(B)=1/3*1/6=1/18

Answer 2

Answer: the answer is 1/18, got it right on the test