GMAT Math : Calculating probability

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Example Questions

Example Question #1 : Data Interpretation

Two dice are rolled. What is the probability that the sum of both dice is greater than 8?

Possible Answers:

$\small \frac{5}{18}$

$\small \frac{5}{9}$

$\small \frac{1}{8}$

$\small \frac{7}{18}$

Correct answer:

$\small \frac{5}{18}$

Explanation:

There are 36 possible outcomes ( $6\times 6=36$ ). 10 out of the 36 outcomes are greater than 8: (6 and 3)(6 and 4)(6 and 5)(6 and 6)(5 and 4)(5 and 5)(5 and 6)(4 and 5)(4 and 6)(3 and 6).

$\small \frac{10}{36}\ =\ \frac{5}{18}$

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Example Question #1 : Data Interpretation

Among a group of 300 people, 15% play soccer, 21% play baseball, and 9% play both soccer and baseball. If one person is randomly selected, what is the probability that the person selected will be one who plays baseball but NOT soccer?

Possible Answers:

$\frac{3}{25}$

$\frac{9}{50}$

$\frac{4}{7}$

$\frac{21}{100}$

Correct answer:

$\frac{3}{25}$

Explanation:

Since there are 300 people, $300\cdot .21=63$ people play baseball and $300\cdot .09=27$ of those people play both baseball and soccer. Therefore, there are 63-27=36 people who play baseball but not soccer.

Probability: $\frac{36}{300}=\frac{3}{25}$

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Example Question #1 : Calculating Probability

If a die is rolled twice, what is the probability that it will land on either 2 or an odd number both times?

Possible Answers:

$\frac{4}{9}$

$\frac{1}{2}$

$\frac{2}{9}$

$\frac{17}{18}$

$\frac{2}{3}$

Correct answer:

$\frac{4}{9}$

Explanation:

probability on one roll: $\frac{4}{6}=\frac{2}{3}$

for both times= $\frac{2}{3}\times \frac{2}{3}= \frac{4}{9}$

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Example Question #2 : Calculating Probability

What is the probability of sequentially drawing 3 aces from a deck or regular playing cards when the selected cards are not replaced?

Possible Answers:

$\frac{1}{2197}$

$\frac{1}{5525}$

$\frac{1}{64}$

$\frac{1}{140,608}$

Correct answer:

$\frac{1}{5525}$

Explanation:

The probability of drawing an ace first is $\frac{4}{52}$ or $\frac{1}{13}$ .

Assuming an ace is the first card selected, the probability of selecting another ace is $\frac{3}{51}$ or $\frac{1}{17}$ .

For the third card, the probability is $\frac{2}{50}$ or $\frac{1}{25}$ .

To calculate the probability of all 3 events happening, you must multiply the probabilities:

$\frac{1}{13}\times\frac{1}{17}\times\frac{1}{25}=\frac{1}{5525}$

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Example Question #5 : Data Interpretation

How many even four-digit numbers larger than 4999 can be formed from the numbers 2, 4, 5, and 7 if each number can be used more than once?

Possible Answers:

256

Correct answer:

Explanation:

Since the number must be larger than 4999, the thousand’s digit has to be 5 or 7. We are also told that the number must be even. Thus, the unit’s digit must be 2 or 4. The middle digits can by any of the numbers 2,4,5, or 7. Therefore, we have a total of $2\cdot 4\cdot 4\cdot 2=64$ possibilities.

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Example Question #6 : Data Interpretation

What is the probability of rolling an even number on a standard dice?

Possible Answers:

0.167

0.5

0.666

0.333

Correct answer:

0.5

Explanation:

A standard dice has 6 faces numbered 1,2,3,4,5,6 .

There are even numbers, 2,4,6 , divided by the total number of faces:

$\frac{3}{6}= 0.5$

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Example Question #7 : Data Interpretation

Shawn is competing in an archery tournament. He gets to shoots three arrows at a target, and his best two shots count.

He hits the bullseye with 40% of his shots. What is the probability that he will hit the bullseye at least twice out of the three times?

Possible Answers:

0.784

0.352

0.192

0.8

0.48

Correct answer:

0.352

Explanation:

There are three scenarios favorable to this event.

1: He hits a bullseye with his first two shots; the third shot doesn't matter.

The probability of this happening is $0.40 \cdot 0.40 = 0.16$

2: He hits a bullseye with his first shot, misses with his second shot, and hits with his third shot.

The probability of this happening is $0.40 \cdot 0.60 \cdot 0.40 = 0.096$

3: He misses with his first shot and hits a bullseye with his other two shots.

The probability of this happening is $0.60 \cdot 0.40 \cdot 0.40 = 0.096$

Add these probabilities:

0.096 + 0.096 + 0.16 = 0.352

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Example Question #195 : Problem Solving Questions

Square

A store uses the above target for a promotion. For each purchase, a customer gets to toss a dart at the target, and the outcome decides his prize. If he hits a pink region, he gets nothing; if he hits a red region, he gets a 10% discount on a future purchase; if he hits a green region, he gets a 20% discount; if he hits a blue region, he gets a 40% discount.

Assume a customer hits the target and no skill is involved. What are the odds against him getting a discount?

Possible Answers:

$1\textrm{ to }1$

$5\textrm{ to }2$

$2\textrm{ to }1$

$3\textrm{ to }1$

$3\textrm{ to }2$

Correct answer:

$1\textrm{ to }1$

Explanation:

The customer gets a discount if he does not hit a pink region. There are ten out of twenty ways to hit a pink region and ten to not hit one - this makes the odds 10 to 10, or, in lowest terms, 1 to 1 against a discount.

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Example Question #5 : Probability

It costs $10 to buy a ticket to a charity raffle in which three prizes are given - the grand prize is $3,000, the second prize is $1,000, and the third prize is $500. Assuming that all of 1,000 tickets are sold, what is the expected value of one ticket to someone who purchases it?

Possible Answers:

$\$4.50$

$-\$5.50$

$\$0$

$-\$4.50$

$\$5.50$

Correct answer:

$-\$5.50$

Explanation:

If 1,000 tickets are sold at $10 apiece, then $10,000 will be raised. The prizes are $3,000, $1,000, and $500, so $4,500 will be given back, meaning that the 1,000 ticket purchasers will collectively lose $5,500. This means that on the average, one ticket will be worth

$\frac{-\$5,500}{1,000} = -\$5.50$

This is the expected value of one ticket.

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Example Question #2 : Data Interpretation

Daria has 5 plates: 2 are green, 1 is blue, 1 is red, and 1 is both green and blue. What is the probability that Daria randomly selects a plate that has blue OR green on it?

Possible Answers:

$\dpi{100} \small \frac{4}{5}$

$\dpi{100} \small \frac{2}{5}$

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

$\dpi{100} \small \frac{3}{5}$

Correct answer:

$\dpi{100} \small \frac{4}{5}$

Explanation:

The easiest way to solve this is by using the complement. Only one of the five plates is NOT blue or green. So $\dpi{100} \small \frac{1}{5}$ of the plates are NOT blue or green. Therefore $\dpi{100} \small 1-\frac{1}{5}=\frac{4}{5}$ of the plates are blue or green.

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GMAT Math : Calculating probability

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #1 : Data Interpretation

Example Question #1 : Data Interpretation

Example Question #1 : Calculating Probability

Example Question #2 : Calculating Probability

Example Question #5 : Data Interpretation

Example Question #6 : Data Interpretation

Example Question #7 : Data Interpretation

Example Question #195 : Problem Solving Questions

Example Question #5 : Probability

Example Question #2 : Data Interpretation

All GMAT Math Resources

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