Probability Question

Colouryellowblackred

Probability1/7??

Daniel has a bag of marbles. he has twice as many black marbles as red marbles. the rest are yellow. he is going to take a marble at random from the bag. yellow is 1/7 what is black and red probability?

P(yellow) = 1/7

so

P(red or black) = 6/7

and there are twice as many black as red so that must be 

P(red)=2/7

P(black)=4/7

n = b + r + y    n = total

b = 2r             twice as many black as red

y/n = 1/7       yellow probability

n = 3r + y

y/(3r + y) = 1/7

7y = 3r + y

6y = 3r

r = 2y

b = 4y

r/n = 2y/n = 2/7    probability of drawing a red

b/n = 4y/n = 4/7   probability of drawing a black

(I see Melody beat me to it.  Luckily, I got the same results as her!).

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Probability Question

Daniel has a bag of marbles. he has twice as many black marbles as red marbles.

the rest are yellow.

he is going to take a marble at random from the bag.

yellow is 1/7

what is black and red probability?

\(\begin{array}{|rcll|} \hline b=2r \\ \hline \end{array}\)

\(\begin{array}{|lcrcll|} \hline \text{Probability yellow } = \frac17 & \Rightarrow & y+b+r &=& 7 & \text{ and } \quad y = 1 \\ & & y+b+r &=& 7 & \qquad b=2r \quad y = 1 \\ & & 1+2r+r &=& 7 \\ & & 3r &=& 6 \\ & & \mathbf{ r } & \mathbf{=} & \mathbf{2} \\ \text{Probability red:} & & \mathbf{ \frac{r}{y+b+r} } & \mathbf{=} & \mathbf{\frac27 } \\\\ & & y+b+r &=& 7 & \qquad r=2 \quad y = 1 \\ & & 1+b+2 &=& 7 & \\ & & b &=& 7-3 & \\ & & \mathbf{ b } & \mathbf{=} & \mathbf{4} \\ \text{Probability black:} & & \mathbf{ \frac{b}{y+b+r} } & \mathbf{=} & \mathbf{\frac47 } \\\\ \hline \end{array} \)

\(\begin{array}{|l|r|r|r|} \hline \text{Colour} & \text{yellow} & \text{black} & \text{red} \\ \hline \text{Probability} & \frac17 & \frac47 & \frac27 \\ \hline \end{array} \)

How lucky!  We all got the same answer :DD