Question

A bo contains 90 discs,numbered from 1 to 90. If one disc is drawn at random from the box,find the probability that it bears (1)a prime number less than 23 (2)a perfect square number

Answer 1

Total outcomes = 90-1=89

a prime number less than 23 are

2, 3, 5, 7, 11, 13, 17, 19

probability of PM.= 8/89



perfect square number are

1, 4, 9, 16, 25, 36, 49, 64, 81, 

probability of P.S.= 9/89

Answer 2

$\underline{\underline{\huge{\textbf{Solution:}}}}$

Given,
A box contains 90 discs numbered from 1 to 90

Experiment - One disc is drawn at random from the box

Let $E_{1}$ be the Event that drawn card bears a $\textit{Prime Number less than 23}$

Let $E_{1}$ be the Event that drawn card bears a $\textit{Perfect square Number}$

$P(E_{1})= ?$

$P(E_{2}) = ?$

$\underline{\large{\textsf{Step-1:}}}}$
Find No. of Total Outcomes when a disc is drawn from the box

No. of Total Outcomes = No. of ways of drawing a disc from box containing 90 discs

$\implies \textsf{No. of Total Outcomes = 90}$

$\underline{\large{\textsf{Step-2:}}}}$
Find the No. of Favorable outcomes for occurrence of Event $E_{1}$ $[n(E_{1})]$

No. of Favorable outcomes for happening of Event $E_{1}$ = No. of Prime Numbers less than 23

$\textsf{ E_{1} = {2, 3, 5, 7, 11, 13, 17, 19}}$

$\implies \textbf{ n(E_{1}) = 8}$

$\underline{\large{\textsf{Step-3:}}}}$
Find the No. of Favorable outcomes for occurrence of Event $E_{2}$ $[n(E_{2})]$

No. of Favorable outcomes for happening of Event $E_{2}$ = No. of Perfect square less than 90

$\textsf{ E_{2} = {1, 4, 9, 16, 25, 36, 49, 64, 81}}$

$\implies \textbf{ n(E_{2}) = 9}$

$\underline{\large{\textsf{Step-4:}}}}$
Find Probability of Occurrence of Events $E_{1}$ and $E_{2}$

We have,
$\bigstar \mathbf{Probability\: of\: Occurrence\: of\: Event = \dfrac{No.\: of\: favorable\: outcomes\: for\: Event}{Total\: No.\: of\: Outcomes\: in\: Experiment}}$

$\implies P(E_{1}) = \dfrac{n(E_{1})}{n(S)}$

Substituting Values

$\implies P(E_{1}) =\dfrac{8}{90}$

$\implies \textbf{ P(E_{1}) = \dfrac{4}{45} }$


$\implies P(E_{2}) = \dfrac{n(E_{2})}{n(S)}$

Substituting Values

$\implies P(E_{2}) =\dfrac{9}{90}$

$\implies \textbf{ P(E_{2}) = \dfrac{1}{10} }$

$\therefore$

$\bigstar \textbf{Probability that drawn card bears a Prime Number less than 23 =\underline{\mathbf{ \dfrac{4}{45} }}}$

$\: \: \: \, \bigstar \textbf{Probability that drawn card bears a Perfect square Number = \underline{\mathbf{ \dfrac{1}{10} }}}$