Question

21. A bag contains 24 balls of which x are red, 2x are white and 3x am are. A ball is selected atrandom. What is the probability that it(i) not red?(ii) white(​

Answer 1

Total Balls in Bag = 24
Red, White and, Blue are x, 2x and 3x
Find the Given Probabilities.
↠ Total Balls = 24
↠ (Red + White + Blue) Balls = 24
↠ (x + 2x + 3x) = 24
↠ 6x = 24
Dividing both term by 6
↠ x = 4

↠ x = 4
\rule{200}{1}
\bigstar \:\boxed{\bf{Probability = \dfrac{Favorable \:Outcomes}{Total \:Outcomes}}}
\rule{100}{2}
\textbf{\textdagger} \: \underline{\large{\frak{Probability \:of \:not\:getting \:red :}}}
\implies\sf P(NR)=\dfrac{n(Not\:Red)}{n(S)}\\\\\implies\sf P(NR)=\dfrac{n(White + Blue)}{n(S)} \\ \\\implies\sf P(NR)=\dfrac{2x + 3x}{24} \\ \\\implies\sf P(NR)=\dfrac{5x}{24} \\ \\\implies\sf P(NR)=\dfrac{5(4)}{24} \\ \\\implies\sf P(NR)= \cancel\dfrac{20}{24} \\ \\\implies \green{\sf P(NR)=\dfrac{5}{6}}
⠀
\therefore \textsf{Probability that it is not red}\: \boxed{\mathbf{P(NR) = \dfrac{5}{6}}}
\rule{200}{2}
\textbf{\textdagger} \: \underline{\large{\frak{Probability \:of \:getting \:white :}}}
\implies\sf P(W)=\dfrac{n(White)}{n(S)} \\ \\\implies\sf P(W)=\dfrac{2x}{24} \\ \\\implies\sf P(W)=\dfrac{2(4)}{24} \\ \\\implies\sf P(W)= \cancel\dfrac{8}{24} \\ \\\implies\green{\sf P(W)=\dfrac{1}{3}}
⠀
\therefore \textsf{Probability that it is white}\: \boxed{\mathbf{P(W) = \dfrac{1}{3}}}