I keep finding myself getting stuck on a conditional probability question.
The question is: "A spinner has 40% of it colored Blue, 30% colored red, and 30% colored green. Suppose that you know that the spinner landed on the same color twice. What is the probability that the color it landed on was red"
I know the rule P(A given B)*P(B)=P(B given A)*P(A), but I can't figure out how I would apply that. Thank you!
On
It depends on how you define the events. Since the spinner landed twice, you can define
P(A): the probability that the last colour it landed was red. 30%
P(B): the probability that the spinner landed on the same colour twice. 34%
P(A|B): the probability that the last colour the spinner landed was red given the spinner landed on the same colour twice. 26.5%
P(B|A): the probability that the spinner landed on the same colour twice given the last colour it landed was red. 30%
So that P(A|B)P(B)=P(B|A)P(A)
You just need to figure out what the events $A$ and $B$ stand for in your problem.
You want to find: $\mathbb{P}(A\mid B)$ where A:"spin color is red" and B:"spin colours are the same".
Leting $S_1$ be the color of spin 1, and $S_2$ be that of spin 2, both of which can take on values: $r$ for red, $b$ for blue, or $g$ for green.
You want to find: $\mathbb{P}(S_1 = r \mid S_1=S_2)$ so it's:
$$\begin{aligned}\mathbb{P}(S_1\!=\!r \mid S_1\!=\!S_2) & =\frac{\mathbb{P}(S_1\!=\!r)\!\cdot\!\mathbb{P}(S_1\!=\!S_2 \mid S_1=r)}{\mathbb{P}(S_1\!=\!S_2)} \\ ~ & = \frac{\mathbb{P}(S_1\!=\!r)\!\cdot\!\mathbb{P}(S_2\!=\!r)}{\mathbb{P}(S_1\!=\!r)\!\cdot\!\mathbb{P}(S_2\!=\!r)+\mathbb{P}(S_1\!=\!b)\!\cdot\!\mathbb{P}(S_2\!=\!b)+\mathbb{P}(S_1\!=\!g)\!\cdot\!\mathbb{P}(S_2\!=\!g)}\\ & = \frac{0.30^2}{0.30^2+0.40^2+0.30^2} \\ ~ & = \frac {0.090}{0.34} \\ ~ & \approx 26.5\ldots\%\end{aligned}$$