A game show features three identical mystery boxes. One box is selected at random for a contestant, who then randomly draws a single coin from the chosen box. Box 1 contains 2 gold coins and 3 silver coins. Box 2 contains 4 gold coins and 1 silver coin. * Box 3 contains 1 gold coin and 4 silver coins. After the contestant draws, it is revealed that the coin is gold. What is the probability that it came from Box 2?

Solution available with step-by-step explanation

SAT Probability Question 51 | Free SAT Prep | Aniko

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Question 51·Hard·Probability and Conditional Probability

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A game show features three identical mystery boxes. One box is selected at random for a contestant, who then randomly draws a single coin from the chosen box.

Box 1 contains 2 gold coins and 3 silver coins.
Box 2 contains 4 gold coins and 1 silver coin.
Box 3 contains 1 gold coin and 4 silver coins.

After the contestant draws, it is revealed that the coin is gold. What is the probability that it came from Box 2?

When you see wording like “Given that the coin is gold, what is the probability it came from Box 2?”, translate it into conditional probability: $P(\text{Box 2} \mid \text{gold}) = \dfrac{P(\text{Box 2 and gold})}{P(\text{gold})}$ . Quickly list: (1) the chance of picking each box, (2) the chance of gold from each box, (3) the total chance of gold by adding the three products, and (4) the joint chance for the specific box and gold. Then take the ratio and simplify the fraction, matching it to the closest answer choice. This structured approach avoids guesswork and keeps multi-step probability problems manageable under time pressure.

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Break the process into stages

First, think of the game in two steps: (1) a box is chosen at random, and then (2) a coin is drawn from that box. How likely is each box to be chosen?

Find gold chances for each box

Independently of the overall question, calculate the probability of drawing a gold coin from Box 1, from Box 2, and from Box 3.

Total probability of gold

Use the fact that each box is chosen with probability $\frac{1}{3}$ to find the overall probability of getting a gold coin from the game, combining the contributions from all three boxes.

Use conditional probability

To find the probability that the coin came from Box 2 given that it is gold, use the idea $P(\text{Box 2} \mid \text{gold}) = \dfrac{P(\text{Box 2 and gold})}{P(\text{gold})}$ .

Desmos Guide

Compute the total probability of drawing a gold coin

In Desmos, enter the expression (1/3)*(2/5) + (1/3)*(4/5) + (1/3)*(1/5) on a line. This represents the total probability of getting a gold coin from any box. Note the value that Desmos gives for this expression.

Compute the joint probability of choosing Box 2 and getting gold

On the next line, enter (1/3)*(4/5). This represents the probability of both choosing Box 2 and drawing a gold coin from it. Again, note the value of this expression.

Form the conditional probability in Desmos

Now enter ( (1/3)*(4/5) ) / ( (1/3)*(2/5) + (1/3)*(4/5) + (1/3)*(1/5) ). The numerical value Desmos returns is the probability that the gold coin came from Box 2; compare this value with the answer choices.

Step-by-step Explanation

Identify what probability is being asked

We are asked: Given that the drawn coin is gold, what is the probability it came from Box 2?

In probability notation, this is $P(\text{Box 2} \mid \text{gold})$ , which means a conditional probability: the probability of Box 2 given that the coin is gold.

A useful idea is:

P(\text{Box 2} \mid \text{gold}) = \frac{P(\text{Box 2 and gold})}{P(\text{gold})}

So we need two things:

$P(\text{Box 2 and gold})$
$P(\text{gold})$ (from any box)

Find the probability of drawing a gold coin from each box

First, find the probability of a gold coin if a specific box is already chosen:

Box 1: 2 gold, 3 silver, total 5 coins
- $P(\text{gold} \mid \text{Box 1}) = \frac{2}{5}$
Box 2: 4 gold, 1 silver, total 5 coins
- $P(\text{gold} \mid \text{Box 2}) = \frac{4}{5}$
Box 3: 1 gold, 4 silver, total 5 coins
- $P(\text{gold} \mid \text{Box 3}) = \frac{1}{5}$

Also, the contestant is equally likely to pick any box, so:

$P(\text{Box 1}) = \frac{1}{3}$
$P(\text{Box 2}) = \frac{1}{3}$
$P(\text{Box 3}) = \frac{1}{3}$

Compute the total probability of getting a gold coin

Use a "weighted sum" over the three boxes:

P(\text{gold}) = P(\text{Box 1})\cdot P(\text{gold} \mid \text{Box 1}) + P(\text{Box 2})\cdot P(\text{gold} \mid \text{Box 2}) + P(\text{Box 3})\cdot P(\text{gold} \mid \text{Box 3})

Substitute the numbers:

P(\text{gold}) = \frac{1}{3}\cdot\frac{2}{5} + \frac{1}{3}\cdot\frac{4}{5} + \frac{1}{3}\cdot\frac{1}{5}

Combine the terms inside the parentheses first:

= \frac{1}{3}\left(\frac{2}{5} + \frac{4}{5} + \frac{1}{5}\right) = \frac{1}{3}\cdot \frac{7}{5} = \frac{7}{15}

So $P(\text{gold}) = \frac{7}{15}$ . Also, the joint probability of choosing Box 2 and getting gold is:

P(\text{Box 2 and gold}) = P(\text{Box 2})\cdot P(\text{gold} \mid \text{Box 2}) = \frac{1}{3}\cdot \frac{4}{5} = \frac{4}{15}

Form the conditional probability and match the answer choice

Now use the conditional probability formula:

P(\text{Box 2} \mid \text{gold}) = \frac{P(\text{Box 2 and gold})}{P(\text{gold})} = \frac{\frac{4}{15}}{\frac{7}{15}}

Dividing two fractions with the same denominator cancels the denominator:

= \frac{4}{7}

So the probability that the gold coin came from Box 2 is $\frac{4}{7}$ , which corresponds to answer choice C.

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Step-by-step Explanation

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