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)=(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)=(4)/(15) \\ P(B)=(8)/(15) \\ P(G)=(3)/(15)=(1)/(5) \end{gathered}`

So

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

Hence, the probability of

(b)

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