In a park there are three picnic areas — A, B, and C. Area A has 4 tables with umbrellas and 8 tables without umbrellas. Area B has 6 tables with umbrellas and 6 tables without umbrellas. * Area C has 7 tables with umbrellas and 2 tables without umbrellas. To assign a table to a visitor, the park manager first chooses one of the three areas at random, each equally likely, and then selects a table at random from the chosen area. Given that the assigned table has an umbrella, what is the probability that it is in Area C?

Solution available with step-by-step explanation

SAT Probability Question 15 | Free SAT Prep | Aniko

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Question 15·Hard·Probability and Conditional Probability

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In a park there are three picnic areas — A, B, and C.

Area A has 4 tables with umbrellas and 8 tables without umbrellas.
Area B has 6 tables with umbrellas and 6 tables without umbrellas.
Area C has 7 tables with umbrellas and 2 tables without umbrellas.

To assign a table to a visitor, the park manager first chooses one of the three areas at random, each equally likely, and then selects a table at random from the chosen area.

Given that the assigned table has an umbrella, what is the probability that it is in Area C?

For conditional probability questions with a two-step process (like choosing a group, then an item), quickly sketch a small table or tree: list each group, its probability of being chosen, and the probability of the desired outcome within that group. Compute two things: (1) the total probability of the given condition happening (here, getting an umbrella) by summing over all groups, and (2) the joint probability of being in the specific group and meeting the condition (here, Area C and umbrella). Then use the formula $P(\text{group} \mid \text{condition}) = P(\text{group and condition}) / P(\text{condition})$ . This structured approach avoids common errors like treating all items as equally likely when the selection is actually made by groups first.

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Identify what probability is being asked

The question gives you that the table has an umbrella and asks for the chance it is in Area C. Think of this as $P(\text{Area C} \mid \text{umbrella})$ .

Break the process into stages

First the manager picks an area (each with probability $1/3$ ), then a table inside that area. Compute the probability of getting an umbrella from A, B, and C separately.

Use the total and joint probabilities

Find the overall probability that a table has an umbrella (combining all three areas). Then find the probability that the manager both chooses Area C and gets an umbrella. Finally, use $P(C \mid \text{umbrella}) = P(C \text{ and umbrella}) / P(\text{umbrella})$ .

Desmos Guide

Compute the overall probability of getting an umbrella

In Desmos, type the expression for $P(\text{umbrella})$ :

(1/3)*(4/12) + (1/3)*(6/12) + (1/3)*(7/9)

Note the decimal (or fraction, if Desmos shows it) output; this is the total probability that a randomly assigned table has an umbrella.

Compute the probability of Area C and umbrella together

In a new line, enter the joint probability:

(1/3)*(7/9)

This is the probability that the manager chooses Area C and then an umbrella table within C.

Form the conditional probability

In another line, enter the ratio:

((1/3)*(7/9)) / ((1/3)*(4/12) + (1/3)*(6/12) + (1/3)*(7/9))

The value Desmos gives for this expression is the probability that the table is in Area C, given that it has an umbrella; match this to the equivalent fraction among the answer choices.

Step-by-step Explanation

Organize the information and the process

The manager:

Chooses Area A, B, or C at random, each with probability $1/3$ .
Then chooses a random table within that area.

Count tables and umbrellas in each area:

Area A: 4 umbrellas, 8 without, so 12 tables total.
Area B: 6 umbrellas, 6 without, so 12 tables total.
Area C: 7 umbrellas, 2 without, so 9 tables total.

Because the areas are chosen first with equal probability, tables are not all equally likely overall (tables in C are slightly more likely than tables in A or B), so we must respect the two-step process in our probabilities.

Find the probability of getting an umbrella from each area

Compute the probability that the table has an umbrella, given each area was chosen:

From A: $P(\text{umbrella} \mid A) = 4/12 = 1/3$ .
From B: $P(\text{umbrella} \mid B) = 6/12 = 1/2$ .
From C: $P(\text{umbrella} \mid C) = 7/9$ .

Now find the overall probability of getting an umbrella by weighting each by $1/3$ (since each area is equally likely):

P(\text{umbrella}) = \frac{1}{3}\cdot\frac{1}{3} + \frac{1}{3}\cdot\frac{1}{2} + \frac{1}{3}\cdot\frac{7}{9}

Convert to a common denominator (54) and add:

P(\text{umbrella}) = \frac{6}{54} + \frac{9}{54} + \frac{14}{54} = \frac{29}{54}

Find the probability of choosing Area C and getting an umbrella

We now want the probability that the manager both:

Chooses Area C, and
Then gets an umbrella in that area.

This is a joint probability:

P(C \text{ and umbrella}) = P(C) \cdot P(\text{umbrella} \mid C)

We know $P(C) = 1/3$ and $P(\text{umbrella} \mid C) = 7/9$ , so:

P(C \text{ and umbrella}) = \frac{1}{3} \cdot \frac{7}{9} = \frac{7}{27}

Use conditional probability to find the final answer

We are asked for the probability that the table is in Area C given that it has an umbrella. This is a conditional probability:

P(C \mid \text{umbrella}) = \frac{P(C \text{ and umbrella})}{P(\text{umbrella})}

Substitute the values we found:

P(C \mid \text{umbrella}) = \frac{\dfrac{7}{27}}{\dfrac{29}{54}}

To divide fractions, multiply by the reciprocal:

P(C \mid \text{umbrella}) = \frac{7}{27} \cdot \frac{54}{29} = \frac{7 \cdot 54}{27 \cdot 29}

Since $54/27 = 2$ , this simplifies to:

P(C \mid \text{umbrella}) = \frac{7 \cdot 2}{29} = \frac{14}{29}

So the correct answer is choice B, $\dfrac{14}{29}$ .

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Step-by-step Explanation

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Strategy

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Step-by-step Explanation