Question

Consider the following set of numbers: 6, 16, 26, 36, 46, 56, 66, 108. If a number is selected at random from this set, what is the probability of selecting a number that is divisible by 6 and 9? *0.1250.250.500.751.00

Answer 1

Given:

The set of numbers are;

`\begin{gathered} 6,16,26,36,46,56,66,108 \\ \text{Total number of outcomes = 8} \end{gathered}`
`\begin{gathered} \text{From the set given, the numbers that can be divisible by 6 are;} \\ 6,36,66,108 \end{gathered}`

Also,

`\begin{gathered} \text{From the set given, the numbers that can be d}ivisible\text{ by 9 are;} \\ 36,108 \end{gathered}`
`\begin{gathered} \text{Probability is given by;} \\ \frac{n\text{ umber of required outcomes}}{n\text{ umber of total outcomes}} \end{gathered}`

`\begin{gathered} N\text{ umbers divisble by 6 = 6,36,66,108} \\ N\text{ umbers divisible by 9 = 36,108} \\ \text{Numbers divisible by 6 and 9 = 36 and 108} \\ \text{Hence, 2 numbers are divisible by 6 and 9 from the set} \\ Total\text{ outcomes from the set = 8} \\ Probability(n\text{ umbers divisible by 6 and 9) =}(2)/(8) \\ P(n\text{ umbers divisible by 6 and 9) =}(1)/(4) \\ P(n\text{ umbers divisible by 6 and 9) =}0.25 \end{gathered}`

Therefore, the probability of selecting a number that is divisible by 6 and 9 is 0.25