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Math Help - Probability Problem

  1. #1
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    Probability Problem

    One jar contains 6 blue marbles, 5 red marbles, 8 white marbles, and 8 goldenrod marbles. A second jar contains 8 blue marbles, 7 red marbles, 8 white marbles, and 6 goldenrod marbles. A marble is picked at random from the first jar and put into the second jar. A marble is then selected at random from the second jar. Find the probability that the marble selected from the second jar is a goldenrod. Give answer as a fraction reduced to lowest terms.

    I think this problem is easier than I think but I keep getting messed up. Help please!
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  2. #2
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    P(G_2 ) = P(B_1 G_2 ) + P(R_1 G_2 ) + P(W_1 G_2 ) + P(G_1 G_2 )
    P(G_2 ) = P(G_2 |B_1 )P(B_1 ) + P(G_2 |R_1 )P(R_1 ) + P(G_2 |W_1 )P(W_1 ) + P(G_2 |G_1 )P(G_1 )
    I will do {P(G_2 |R_1 )P(R_1 )}=\frac{6}{30}\frac{5}{27}
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  3. #3
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    Hello, sweeetcaroline!

    I'll simplify the wording . . .


    One jar contains 8 Gold marbles and 19 Others.
    A second jar contains 6 Gold marbles and 23 Others.
    A marble is picked at random from the first jar and put into the second jar.
    A marble is then selected at random from the second jar.
    Find the probability that the marble selected from the second jar is Gold.
    There are two cases to consider . . .


    [1] The first marble drawn is Gold: . P(\text{1st Gold}) \:=\:\frac{8}{27}
    . . .Then the second jar contains: 7 Gold and 23 Others: . P(\text{2nd Gold}) \:=\:\frac{7}{30}
    . . . P(\text{1st Gold }\wedge\text{ 2nd Gold}) \:=\:\frac{8}{27}\cdot\frac{7}{30} \:=\:\frac{56}{810}


    [2] The first marble drawn is an Other: . P(\text{1st Other}) \:=\:\frac{19}{27}

    . . .Then the second jar contains: 6 Gold and 24 Others: . P(\text{2nd Gold}) \:=\:\frac{6}{30}

    . . . P(\text{1st Other }\wedge\text{ 2nd Gold}) \:=\:\frac{19}{27}\cdot\frac{6}{30} \:=\:\frac{114}{810}


    Therefore: . P(\text{2nd Gold}) \;=\;\frac{56}{810} + \frac{114}{810} \;=\;\frac{170}{810} \;=\;\boxed{\frac{17}{81}}

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