91影视

Spin spinner I twice, then spinner II. What is the probability of getting a 2, followed by a 2 or a 4, followed by Red or Green?

Letting R stands for Red, Y for Yellow, and G for Green, we are asked to look for $P\left(22\mathrm{R}\right)+P\left(22\mathrm{G}\right)+P\left(24\mathrm{R}\right)+P\left(24\mathrm{G}\right)$.

We added them since they are mutually exclusive events.

Listing the possible outcomes from Spinner I twice then II, we have

$S=\{11\mathrm{R},11\mathrm{Y},11\mathrm{G},12\mathrm{R},12\mathrm{Y},12\mathrm{G},13\mathrm{R},13\mathrm{Y},13\mathrm{G},14\mathrm{R},14\mathrm{Y},14\mathrm{G}21\mathrm{R},21\mathrm{Y},21\mathrm{G},22\mathrm{R},22\mathrm{Y},22\mathrm{G},23\mathrm{R},23\mathrm{Y},23\mathrm{G},24\mathrm{R},24\mathrm{Y},24\mathrm{G}31\mathrm{R},31\mathrm{Y},31\mathrm{G},32\mathrm{R},32\mathrm{Y},32\mathrm{G},33\mathrm{R},33\mathrm{Y},33\mathrm{G},34\mathrm{R},34\mathrm{Y},34\mathrm{G}41\mathrm{R},41\mathrm{Y},41\mathrm{G},42\mathrm{R},42\mathrm{Y},42\mathrm{G},43\mathrm{R},43\mathrm{Y},43\mathrm{G},44\mathrm{R},44\mathrm{Y},44\mathrm{G}\}$

From this , we were able to get the following

$P\left(22R\right)=\frac{1}{48},P\left(22G\right)=\frac{1}{48},P\left(24R\right)=\frac{1}{48},P\left(24G\right)=\frac{1}{48}$

Since they just appeared once in the sample space with a total of 48 possible outcomes.

Thus,$P\left(22\mathrm{R}\right)+P\left(22\mathrm{G}\right)+P\left(24\mathrm{R}\right)+P\left(24\mathrm{G}\right)=\frac{1}{48}+\frac{1}{48}+\frac{1}{48}+\frac{1}{48}=\frac{4}{48}=\frac{1}{12}$.

The possible outcome is

$S=\{11\mathrm{R},11\mathrm{Y},11\mathrm{G},12\mathrm{R},12\mathrm{Y},12\mathrm{G},13\mathrm{R},13\mathrm{Y},13\mathrm{G},14\mathrm{R},14\mathrm{Y},14\mathrm{G}21\mathrm{R},21\mathrm{Y},21\mathrm{G},22\mathrm{R},22\mathrm{Y},22\mathrm{G},23\mathrm{R},23\mathrm{Y},23\mathrm{G},24\mathrm{R},24\mathrm{Y},24\mathrm{G}31\mathrm{R},31\mathrm{Y},31\mathrm{G},32\mathrm{R},32\mathrm{Y},32\mathrm{G},33\mathrm{R},33\mathrm{Y},33\mathrm{G},34\mathrm{R},34\mathrm{Y},34\mathrm{G}41\mathrm{R},41\mathrm{Y},41\mathrm{G},42\mathrm{R},42\mathrm{Y},42\mathrm{G},43\mathrm{R},43\mathrm{Y},43\mathrm{G},44\mathrm{R},44\mathrm{Y},44\mathrm{G}\}$

The probability is$\frac{1}{12}$.