Question

Twenty counters are placed in a bag. Eight of them are red, six are yellow, four are green and the remaining blue. If one counter is picked at random, what is the probability that it is any color except green?

Answer 1

Step 1 Finding the total number of outcomes and the number of favorable outcomes:

The event is selecting a counter at random.

There are $20$ counters placed in a bag.

So, the total number of outcomes is $20$.

Out of the total counters, $8$ are red, $6$ are yellow, $4$ are green and the remaining are blue.

$$\begin{gathered} Anycolorexceptgreen=Totalcounters-Greencounters \\ Anycolorexceptgreen=20-4 \\ Anycolorexceptgreen=16 \end{gathered}$$

There are $16$ any color except green counters in the bag.

So, the number of favorable outcomes is $16$.

Step 2 Finding the probability that the counter is any color except green:

The formula for probability is,

$$\begin{gathered} \Rightarrow Probability=\frac{Numberoffavorableoutcomes}{Totalnumberofoutcomes} \\ \Rightarrow P(anycolorexceptgreen)=\frac{16}{20} \\ \therefore P(anycolorexceptgreen)=\frac{4}{5} \end{gathered}$$

Therefore, the probability of any color except green counters is $\frac{4}{5}$.