Three spinners are marked with equal amounts of Red, Blue and Yellow. At a particular instance, all three are spun together. What is the probability that at least two of the spinners land on red?
The at least part is confusing me.
My attempt: So if all three lands on red the probability will be: $P(All\space red)=\frac{1}{3}\times\frac{1}{3}\times\frac{1}{3}=\frac{1}{27}$ Having at least two will have a higher probability; two of the spinners should have red and the other can have any colour:$$P(Two\space red\space and\space one\space any)=\frac{1}{3}\times\frac{1}{3}\times\frac{3}{3}=\frac{1}{9}$$
$\endgroup$2 Answers
$\begingroup$i) if two of them land on red - choose which two spinners land on red, the third can be either blue or yellow. So if $X$ is the event of a spinner landing on red,
$ \displaystyle P(X=2) = {3 \choose 2} \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{2}{3} = \frac{2}{9}$
ii) All spinners land on red,
$ \displaystyle P(X=3) = \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{1}{3} = \frac{1}{27}$
Adding them,
$\displaystyle P(X \geq 2) = \frac{7}{27}$
$\endgroup$ 1 $\begingroup$In total there are $27$ possible spins, all of which have equal probability. In one spin, all three land on red. In $6$ spins, two land on red while the third lands on either blue or yellow. So the probability that at least two land on red in $\frac{7}{27}.$
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