GMAT Math : Data Interpretation

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Example Question #191 : Word Problems

A die is rolled and then a coin is tossed. What is the probability that the die shows an even number AND the coin shows a tail?

Possible Answers:

$\dpi{100} \small \frac{1}{3}$

$\dpi{100} \small \frac{1}{2}$

$\dpi{100} \small \frac{1}{4}$

$\dpi{100} \small \frac{1}{5}$

$\dpi{100} \small \frac{1}{6}$

Correct answer:

$\dpi{100} \small \frac{1}{4}$

Explanation:

We can calculate the two individual probabilities first.

Prob(die shows even) $\dpi{100} \small =\frac{3}{6}=\frac{1}{2}$ (2, 4, and 6 out of 1, 2, 3, 4, 5, and 6)

Prob(tail) $\dpi{100} \small =\frac{1}{2}$ (tail out of head and tail)

Then, Prob(even AND tail) is $\dpi{100} \small P(even)\times P(tail)=\frac{1}{2}\times \frac{1}{2}$ .

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Example Question #12 : Probability

A jar contains 8 blue marbles and 4 red marbles. What is the probability of picking a blue marble followed by a red marble if the first marble chosen is not put back in the jar?

Possible Answers:

$\frac{23}{66}$

$\frac{2}{11}$

$\frac{8}{33}$

$\frac{3}{4}$

$\frac{2}{9}$

Correct answer:

$\frac{8}{33}$

Explanation:

There are 12 marbles total. The probability of picking a blue marble first is $\frac{8}{12}=\frac{2}{3}$ . The probability of then picking a red marble out of the 11 remaining marbles is $\frac{4}{11}$ . Therefore, the probability is $\frac{2}{3} \times\frac{4}{11}=\frac{8}{33}$ .

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Example Question #13 : Probability

Square

Refer to the above figure, which shows a target. Each of the squares is of equal size. If a dart is thrown at the target, what are the odds against hitting a red region?

You may assume that the dart hits the target, and you may disregard any skill factor.

Possible Answers:

$7\textrm{ to }2$

$3\textrm{ to }1$

$10 \textrm{ to }3$

$20\textrm{ to }3$

$4\textrm{ to }1$

Correct answer:

$3\textrm{ to }1$

Explanation:

There are fifteen ways to not hit a red region, and five ways to hit a red region. This makes the odds 15 to 5, or, in lowest terms, 3 to 1, against hitting a red region with a randomly thrown dart.

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Example Question #14 : Probability

Sheryl is competing in an archery tournament. She gets to shoot three arrows at a target, and her best one counts.

Sheryl hits the bullseye 42% of the time. What is the probability (two decimal places) that she will hit the bullseye at least once in her three tries?

Possible Answers:

0.80

0.07

0.42

0.22

Correct answer:

0.80

Explanation:

This is most easily solved by finding the probability that she will not hit the bullseye at all in her three tries. If she hits 42% of the time, she misses 58% of the time, and the probability she misses three times will be

$0.58 ^{3} = 0.195112 \approx 0.20$ .

The probability of hitting the bullseye at least once in three tries is the complement of this, or 1 - 0.20 = 0.80 .

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Example Question #15 : Probability

If you have a bag with 15 grey marbles, 15 yellow marbles and 20 green marbles, what is the probability of choosing a green marble followed by a yellow marble? Assume no replacement.

Possible Answers:

$\small \frac{3}{25}$

$\small \small \frac{3}{49}$

$\small \small \frac{2}{5}$

$\small \small \frac{3}{50}$

$\frac{6}{49}$

Correct answer:

$\frac{6}{49}$

Explanation:

To calculate probability of multiple events, find the probability of each event and multiply them together. We begin with 50 total marbles in the bag.

For the first case, we are asked to pick a green marble. Because there are 20 green marbles and 50 total, our probability of this event is as follows:

$\small \frac{20}{50}=\frac{2}{5}$

For the second case, we need to pick a yellow marble. However, this time there are only 49 marbles left in the bag, because we didn't replace the green marble. Because we have 15 yellow marbles, the probability of this event is as follows:

$\small \frac{15}{49}$

To find our final answer, we need to multiply these two together.

$\small \frac{2}{5}*\frac{15}{49}=\frac{2}{1}*\frac{3}{49}=\frac{6}{49}$

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Example Question #201 : Gmat Quantitative Reasoning

If you flip a quarter three times, what is the probability of it landing on heads all three times?

Possible Answers:

$\frac{1}{8}$

$\frac{1}{3}$

$\frac{1}{2}$

$\frac{1}{4}$

$\frac{1}{16}$

Correct answer:

$\frac{1}{8}$

Explanation:

Every time you flip a coin, there is a 1 in 2 chance of it landing on heads. So, if we want to know the probability of a coin landing on heads a certain number of times times in a row, we multiply the probability of that occurrence for however many times the coin is flipped. For three flips, this gives us:

$P=\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}=\frac{1}{8}$

So there is a 1 in 8 chance that a coin will land on heads three times in a row.

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Example Question #11 : Probability

A drawer has 4 green socks, 6 blue socks, 12 white socks, 8 black socks, and 2 pink socks. If you reach in and pull out a sock at random, what is the probability the sock will be blue?

Possible Answers:

$\frac{3}{16}$

$\frac{3}{8}$

$\frac{1}{6}$

$\frac{5}{16}$

$\frac{3}{32}$

Correct answer:

$\frac{3}{16}$

Explanation:

If you reach in and pull out a sock at random, the probability of it being blue is equal to the number of blue socks divided by the total number of socks in the drawer. First we'll calculate the total number of socks in the drawer:

$4 \:green + 6 \:blue + 12 \:white + 8 \:black + 2 \:pink = 32 \:socks$

Now we divide the number of blue socks by the total number of socks to find the probability of picking a blue sock:

$P=\frac{6}{32}=\frac{3}{16}$

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Example Question #1 : Diagrams

In a room filled with 100 people, 52 are men and 40 are blonde. There were 16 blonde men in the room. How many non-blonde women were there? Use the following diagram.

Possible Answers:

Correct answer:

Explanation:

Of the 100 people, $\left ( 36+16+24 \right )$ were either men or blonde or both. Therefore, are non-blonde and non-men (women).

100-76=24

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Example Question #1 : Understanding Diagrams

Graph

The above circle graph shows the results of a school election. According to the rules, the students who finish first, second, and third become President, Vice-President, and Secretary-Treasurer respectively. In the event of any tie, a runoff election will be held.

Who was elected Secretary-Treasurer?

Possible Answers:

Crieghton

There will be a runoff between Wells and Hawley

There will be a runoff between Crieghton and Wells

Wells

Hawley

Correct answer:

Wells

Explanation:

The third-largest portion of the graph is the gray portion, which represents Wells. Wells won the office of Secretary-Treasurer outright.

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Example Question #11 : Data Interpretation

$\begin{matrix} \textrm{Crane}&124 \\ \textrm{Jones}&\; 87\\\textrm{Smith} & \; 87\\ \textrm{Trask} & \; 98\\ \textrm{Wells} & \; 75 \end{matrix}$

The above table shows the results of a school election. According to the rules, the students who finish first, second, and third become President, Vice-President, and Secretary-Treasurer respectively. In the event of any tie, a runoff election will be held.

Who was elected Secretary-Treasurer?

Possible Answers:

Jones and Wells will have a runoff to determine who will be Secretary-Treasurer.

Smith and Wells will have a runoff to determine who will be Secretary-Treasurer.

Jones and Smith will have a runoff to determine who will be Secretary-Treasurer.

Smith

Jones

Correct answer:

Jones and Smith will have a runoff to determine who will be Secretary-Treasurer.

Explanation:

Jones and Smith tied for third (87 each), so by the rules, there will be a runoff election for the office of Secretary-Treasurer between these two.

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GMAT Math : Data Interpretation

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #191 : Word Problems

Example Question #12 : Probability

Example Question #13 : Probability

Example Question #14 : Probability

Example Question #15 : Probability

Example Question #201 : Gmat Quantitative Reasoning

Example Question #11 : Probability

Example Question #1 : Diagrams

Example Question #1 : Understanding Diagrams

Example Question #11 : Data Interpretation

All GMAT Math Resources

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