Question

An urn contains 10 red balls, 6 green balls, 15 orange balls, and 14 blue balls. If one ball is randomly drawn from the urn, what are the odds against the ball being blue?State your answer as a ratio (two numbers separated by a colon), but do not put any spaces in your response.

Answer 1

Given:

Red balls =10

Green balls =6

Orange ball =15

Blue ball =14

Total number of balls = 10+6+15+14 = 45

The probability that the randomly drawn ball is blue,

$\begin{gathered} P(B)=\frac{\text{Number of blue balls}}{\text{Total balls}} \\ P(B)=\frac{14}{45} \end{gathered}$

The probability that the randomly drawn ball is not blue,

$\begin{gathered} P(\text{ not B)=1-P(B)} \\ =1-\frac{14}{45}=\frac{31}{45} \end{gathered}$

The odds against event is calculated as,

$\begin{gathered} \text{Odds against=}\frac{P(\text{ not B)}}{P(B)} \\ =\frac{\frac{31}{45}}{\frac{14}{45}} \\ =\frac{31}{45}\times\frac{45}{14} \\ =\frac{31}{14} \end{gathered}$

Answer: the odds against the ball being blue is 31 : 14.