Question

A number is selected from the set {1, 2, 3, 5, 15, 21, 29, 38, 500}. If equal elemental probabilities are assigned, what is the probability that the number chosen is either less than 29 or odd? 6/9 7/9 8/9

Answer 1

The required probability is
`(7)/(9)`.

Explanation:

Set given is:

S = {1, 2, 3, 5, 15, 21, 29, 38, 500}

Total number of elements in set,
`n(S)` = 9

Let A be the event that the number is less than 29 ({1, 2, 3, 5, 15, 21}).

Number of items in the event A,
`n(A)` = 6

Probability of event A,

`P(A) = (n(A))/(n(S))}=(6)/(9) \Rightarrow (2)/(3)`

Formula for probability of any event E:

`P(E) = \frac{\text{Number of favorable cases}}{\text {Total number of cases}}`

Let B be the event that the number is odd (either of {1,3,5,15,21,29}).

Number of items in the event B,
`n(B)` = 6

Probability of event B,

`P(B) = (n(B))/(n(S))}=(6)/(9) \Rightarrow (2)/(3)`

The event A and B have a few elements in common, i.e. numbers less than 29 which are odd as well.

The common elements are represented as:

`A \cap B = \{1, 3, 5, 15, 21\}`

`n(A\cap B) = 5`

`P(A \cap B ) = (n(A \cap B))/(n(s))\\\Rightarrow P(A \cap B) = (5)/(9)`

To find probability of selecting a number which is either less than 29 (event A) or odd (event B),

We have to find
`P(A\ or \ B)` which is represented as
`P(A \cup B)` and the formula is:

`P(A \cup B) = P(A) + P(B) - P(A \cap B)\\\Rightarrow (2)/(3) + (2)/(3) - (5)/(9)\\\Rightarrow (12-5)/(9)\\\Rightarrow (7)/(9)`

The required probability is
`(7)/(9)`.