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

Answer:

The required probability is $\frac{7}{9}$.

Step-by-step 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,

$P(A) = \dfrac{n(A)}{n(S)}}=\dfrac{6}{9} \Rightarrow \dfrac{2}{3}$

Formula for probability of any event E:

$P(E) = \dfrac{\text{Number of favorable cases}}{\text {Total number of cases}}$

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,

$P(B) = \dfrac{n(B)}{n(S)}}=\dfrac{6}{9} \Rightarrow \dfrac{2}{3}$

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:

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

$n(A\cap B) = 5$

$P(A \cap B ) = \dfrac{n(A \cap B)}{n(s)}\\\Rightarrow P(A \cap B) = \dfrac{5}{9}$

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:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)\\\Rightarrow \dfrac{2}{3} + \dfrac{2}{3} - \dfrac{5}{9}\\\Rightarrow \dfrac{12-5}{9}\\\Rightarrow \dfrac{7}{9}$

The required probability is $\frac{7}{9}$.