Math  /  Data & Statistics

Question5. We have two boxes. The first box contains five green marbles and 11 red marbles; the second contains six green marbles and 10 red marbles. Jonathan selects a marble by first choosing one of the two boxes (at random). He then selects one of the marbles in that box (at random). If Jonathan selected a red marble, what is the probability that he selected a ball from the second box?

Studdy Solution

STEP 1

What is this asking? If we know Jonathan picked a red marble, what are the chances he picked it from the second box? Watch out! Don't mix up the probability of picking a red marble *from* the second box with the probability of picking from the second box *given* that we picked a red marble.

STEP 2

1. Probability of picking the first box
2. Probability of picking a red marble from the first box
3. Probability of picking the second box
4. Probability of picking a red marble from the second box
5. Probability of picking a red marble overall
6. Probability of picking from the second box given a red marble

STEP 3

Since there are two boxes and he chooses at random, the probability of picking the first box is 12\frac{1}{2}.
Easy peasy!

STEP 4

The first box has **5** green marbles and **11** red marbles, for a total of 5+11=165 + 11 = 16 marbles.
So, the probability of picking a red marble *if* he chose the first box is 1116\frac{11}{16}.

STEP 5

Just like the first box, the probability of picking the second box is 12\frac{1}{2}.

STEP 6

The second box has **6** green marbles and **10** red marbles, totaling 6+10=166 + 10 = 16 marbles.
The probability of picking a red marble *if* he chose the second box is 1016\frac{10}{16}, which simplifies to 58\frac{5}{8}.

STEP 7

To get the overall probability of picking a red marble, we need to consider both boxes.
We multiply the probability of picking each box by the probability of picking a red marble from that box, and then add those probabilities together.
So, we have: (121116)+(121016)=1132+1032=2132(\frac{1}{2} \cdot \frac{11}{16}) + (\frac{1}{2} \cdot \frac{10}{16}) = \frac{11}{32} + \frac{10}{32} = \frac{21}{32}.
This is the probability of picking a red marble, regardless of which box it came from.

STEP 8

Now, we want the probability that Jonathan chose from the second box *given* that he picked a red marble.
This is where things get interesting!
We're going to use our result from calculating the overall probability of picking a red marble.
We take the probability of picking a red marble from the second box (which is the probability of picking the second box multiplied by the probability of picking a red marble from that box: 121016=1032\frac{1}{2} \cdot \frac{10}{16} = \frac{10}{32}) and divide it by the overall probability of picking a red marble, which we just calculated as 2132\frac{21}{32}.
So, we have 10/3221/32=10323221=1021\frac{10/32}{21/32} = \frac{10}{32} \cdot \frac{32}{21} = \frac{10}{21}.

STEP 9

The probability that Jonathan selected a marble from the second box, given that he selected a red marble, is 1021\frac{10}{21}.

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