Question

There are eight marbles in a bag. Four marbles are blue (B), two marbles are red (R) and two marbles are green(G) Steve takes a marble at random from the bag. (i) What is the probability that Steve will take a blue marble. (ii) What is the probability that Steve will take a green marble. (iii) What is the probability that Steve will take a yellow marble.

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Answer 1

Step-by-step explanation:

probability=no. of favourable outcomes/ total outcomes

total marbles =8

(i) probability= 4/8 =1/2

(ii) probability=2/8=1/4

(iii) probability=0 because there is no yellow marble In the bag

Answer 2

Given

$\bold{Total \ number \ of \ marbles} = 8$

$\bold{Total \ number \ of \ blue \ marbles} = 4$

$\bold{Total \ number \ of \ red \ marbles} = 2$

$\bold{Total \ number \ of \ green \ marbles} = 2$

SOMETING YOU NEED TO KNOW

$\boxed{\sf Probability \ of \ getting \ an \ outcome = \dfrac{Given \ outcome}{Total \ number \ of \ outcomes} }$

$(i) \bold{Probability \ of \ getting \ a \ blue \ marble}$

$\implies \sf \dfrac{Total \ number \ of \ blue \ marbles}{Total \ number \ of \ marbles}$

$= \dfrac{4}{8}$

$=\boxed{\boxed{\bold{\dfrac{1}{2} }}}}}}}}}$

                                                               

$(ii) \bold{Probability \ of \ getting \ a \ green \ marble}$

$\implies \sf \dfrac{Total \ number \ of \ green \ marbles}{Total \ number \ of \ marbles}$

$= \dfrac{2}{8}$

$=\boxed{\boxed{\bold{\dfrac{1}{4} }}}}}}}}}$

                                                                     

$(iii) \bold{Probability \ of \ getting \ a \ yellow \ marble}$

$\implies \sf \dfrac{Total \ number \ of \ yellow \ marbles}{Total \ number \ of \ marbles}$

$= \dfrac{0}{8}$

$=\boxed{\boxed{\bold{0 }}}}}}}}}$