Question

A bag contains 4 white counters, 8 black counters, and 3 green counters. What is the probability of drawing(a) a white counter or a green counter? (b) a black counter or a green counter? (c) not a green counter?

Answer 1

Given:

$\begin{gathered} White-counters(W)=4 \\ Black-counters(B)=8 \\ Green-counters(G)=3 \end{gathered}$

To Determine: The probability of drawing a white or a green counter

Solution

$\begin{gathered} P(A)=\frac{n(A)}{n(S)} \\ P(A)=The\text{ probability of A} \\ n(A)=Number\text{ of A} \\ n(S)=Total\text{ Outcome} \end{gathered}$$\begin{gathered} For\text{ the given} \\ n(S)=4+8+3=15 \\ P(W)=\frac{4}{15} \\ P(B)=\frac{8}{15} \\ P(G)=\frac{3}{15}=\frac{1}{5} \end{gathered}$

So

$\begin{gathered} P(WORG)=P(W)+P(G) \\ =\frac{8}{15}+\frac{1}{5} \\ =\frac{8+3}{15} \\ =\frac{11}{15} \end{gathered}$

Hence, the probability of

(b)

$\begin{gathered} P(BORG)=P(B)+P(G) \\ =\frac{4}{15}+\frac{1}{5} \\ =\frac{4+3}{15} \\ =\frac{7}{15} \end{gathered}$