Question

A box contains 5 red marbles, 8 white marbles and 4 green marbles a box contains 5 red marbles 8 white marble and 4 green marble .One marble taken out of the box at random. what is the probability that the marble taken out. will be(1) red ? (2) white? (3)not green?. ​

Answer 1

FORUMLA = probability=no.of fefavourable outcome/total no. of outcome

Step-by-step explanation:

probability of getting one red ball =5/17

probability of getting two white ball = 8/17

probability of getting Not 3 Green balls = 17/3

Answer 2

${\huge{\boxed{\sf{\green{❥✰Question✰}}}}}$

A box contains 5 red marbles, 8 white marbles and 4 green marbles a box contains 5 red marbles 8 white marble and 4 green marble .One marble taken out of the box at random. what is the probability that the marble taken out. will be(1) red ? (2) white? (3)not green?. ​

${\huge{\underline{\bf{\pink{❥✰➵ANSWER :-✰}}}}}$

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

$\underline{\underline{\red{\frak{To\:find\::}}}}$$\qquad\qquad\:\:\:\bullet\:{\small{\sf {Probability\:of\:drawing\:a\:non\:-\:white\:marble}}}∙$

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

$:\mapsto\:\sf Total\:no\:of\:marbles\:=\:6\:+\:4\:+\:2:↦Totalnoofmarbles=6+4+2:\mapsto\:\sf Total\:no\:of\:marbles\:=\:\bf{12}:↦$

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

Probability=

Total no.of outcomes

Favourable out comes

$★\begin{gathered}\qquad \qquad\tiny \ddag \: {\underline{\frak{Putting\:all\;values\;in\;the\:formula\::-}}} \\\end{gathered} ‡$

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

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

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

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Thanks!

`TheEmeraldBoyy!