Question

A bag has 8 orange balls, 4 green balls and 5 blue balls. Find the probability that the ball chosen is

(i) either an orange or green ball

(ii) Neither orange or green ball

(iii) Orange or green or blue

• A. (i) $\frac{12}{17}$ (ii) $\frac{5}{17}$ (iii) 1
• B. (i) $\frac{8}{17}$ (ii) $\frac{7}{17}$ (iii) 2
• C. (i) $\frac{12}{17}$ (ii) $\frac{5}{17}$ (iii) 1
• D. (i) $\frac{5}{17}$ (ii) $\frac{2}{17}$ (iii) 0

Answer 1

The correct option is A

(i) $\frac{12}{17}$ (ii) $\frac{5}{17}$ (iii) 1

Total number of balls
= Number of orange balls + Number of green balls + Number of blue balls = 8 + 4 + 5 = 17
Number of orange or green ball = 8 + 4 = 12
(i) $\Rightarrow$ P (either an orange or green ball ) = $\frac{12}{17}$

(ii) Neither Orange of green ball

$\Rightarrow$ 1 - P ( (either an orange or green ball ) = 1 -$\frac{12}{17}$

= $\frac{5}{17}$

(iii) Orange or green or blue

$\Rightarrow$ P ( (either an orange or green or blue ball ) = $\frac{17}{17}$

= $\frac{17}{17}$

= 1