Question

there are six red marbles 4 green and 2 white marbles in a box what is the probability of drawing a non white marble​

Answer 1

Given :

There are six red marbles 4 green and 2 white marbles in a box.

To find :

What is the probability of drawing a non white marble ?

Solution :

Total number of marbles = Red marbles + Green marbles + White marbles = 6 + 4 + 2 = 12

• As we know that,

★ Probability = Favourable outcome/Total number of outcome

Consider probability of getting non white be P(E)

• Non white marble = Green marble + red marble = 6 + 4 = 10

→ P(E) = 10/12 = 5/6

•°• The probability of getting non white marble is 5/6

Answer 2

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$\underline{\underline{\red{\frak{Given}}}}\begin{cases} & \sf{\bullet\:No\:of\;red\;marbles\:in\:a\:box\: =\: \bf{6}} \\ & \sf{\bullet\:No\:of\;green\;marbles\:in\:a\:box\: =\: \bf{4}} \\ & \sf{\bullet\:No\:of\;white\;marbles\:in\:a\:box\: =\: \bf{2}} \end{cases}$

$\underline{\underline{\red{\frak{To\:find\::}}}}$

$\qquad\qquad\:\:\:\bullet\:{\small{\sf {Probability\:of\:drawing\:a\:non\:-\:white\:marble}}}$

$\underline{\underline{\red{\frak{Solution\::}}}}$

$:\mapsto\:\sf Total\:no\:of\:marbles\:=\:6\:+\:4\:+\:2$

$:\mapsto\:\sf Total\:no\:of\:marbles\:=\:\bf{12}$

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$:\rightarrow\:\sf No\:of\:non\:-\:white\:marbles\:=\:6\:+\:4$

$:\rightarrow\:\sf No\:of\:non\:-\:white\:marbles\:=\:\bf{10}$

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$\boxed{\boxed{\sf{Probability\:=\:\green{\dfrac{Favourable\:outcomes}{Total\:no.\:of\:outcomes}}}}}\:\red\bigstar$

$\qquad \qquad\tiny \ddag \: {\underline{\frak{Putting\:all\;values\;in\;the\:formula\::-}}}$

$:\implies\:{\small{\sf {P(E)\:of\:drawing\:a\:non\:-\:white\:marble\:=\:\dfrac{10}{12}}}}$

$:\implies\:{\small{\sf {P(E)\:of\:drawing\:a\:non\:-\:white\:marble\:=\:\dfrac{\cancel{10}}{\cancel{12}}}}}$

$:\implies\:{\small{\sf {P(E)\:of\:drawing\:a\:non\:-\:white\:marble\:=\:\dfrac{5}{6}}}}$

$\underline{\boxed {\small {\frak {\therefore \green {P(E)\:of\:drawing\:a\:non\:-\:white\:marble\:\leadsto\: {\red {\dfrac{5}{6}}}}}}}}$

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