Calculating probability - GMAT Quantitative
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What is the probability of rolling an even number on a standard dice?
What is the probability of rolling an even number on a standard dice?
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A standard dice has 6 faces numbered 
.
There are 
even numbers, 
, divided by the total number of faces:
A standard dice has 6 faces numbered
There are
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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?
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?
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We can calculate the two individual probabilities first.
Prob(die shows even) =\frac{3}{6}$=\frac{1}{2}$ (2, 4, and 6 out of 1, 2, 3, 4, 5, and 6)
Prob(tail) =\frac{1}{2}$ (tail out of head and tail)
Then, Prob(even AND tail) is P(even)times P(tail)=\frac{1}{2}$times $\frac{1}{2}$.
We can calculate the two individual probabilities first.
Prob(die shows even) =\frac{3}{6}$=\frac{1}{2}$ (2, 4, and 6 out of 1, 2, 3, 4, 5, and 6)
Prob(tail) =\frac{1}{2}$ (tail out of head and tail)
Then, Prob(even AND tail) is P(even)times P(tail)=\frac{1}{2}$times $\frac{1}{2}$.
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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?
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?
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The easiest way to solve this is by using the complement. Only one of the five plates is NOT blue or green. So $\frac{1}{5}$ of the plates are NOT blue or green. Therefore 1-$\frac{1}{5}$=\frac{4}{5}$ of the plates are blue or green.
The easiest way to solve this is by using the complement. Only one of the five plates is NOT blue or green. So $\frac{1}{5}$ of the plates are NOT blue or green. Therefore 1-$\frac{1}{5}$=\frac{4}{5}$ of the plates are blue or green.
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Two dice are rolled. What is the probability that the sum of both dice is greater than 8?
Two dice are rolled. What is the probability that the sum of both dice is greater than 8?
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There are 36 possible outcomes (
). 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).
$\frac{10}{36}$ = $\frac{5}{18}$
There are 36 possible outcomes (
$\frac{10}{36}$ = $\frac{5}{18}$
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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?
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?
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Since there are 300 people, 
people play baseball and 
of those people play both baseball and soccer. Therefore, there are 
people who play baseball but not soccer.
Probability: 
Since there are 300 people,
Probability:
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If a die is rolled twice, what is the probability that it will land on either 2 or an odd number both times?
If a die is rolled twice, what is the probability that it will land on either 2 or an odd number both times?
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probability on one roll: 
for both times= 
probability on one roll:
for both times=
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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?
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?
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There are 12 marbles total. The probability of picking a blue marble first is 
. The probability of then picking a red marble out of the 11 remaining marbles is 
. Therefore, the probability is 
.
There are 12 marbles total. The probability of picking a blue marble first is
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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?
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?
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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 
possibilities.
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
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What is the probability of sequentially drawing 3 aces from a deck or regular playing cards when the selected cards are not replaced?
What is the probability of sequentially drawing 3 aces from a deck or regular playing cards when the selected cards are not replaced?
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The probability of drawing an ace first is 
or 
.
Assuming an ace is the first card selected, the probability of selecting another ace is 
or 
.
For the third card, the probability is 
or 
.
To calculate the probability of all 3 events happening, you must multiply the probabilities:
The probability of drawing an ace first is
Assuming an ace is the first card selected, the probability of selecting another ace is
For the third card, the probability is
To calculate the probability of all 3 events happening, you must multiply the probabilities:
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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.
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.
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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.
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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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?
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?
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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.
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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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?
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?
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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 
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 
3: He misses with his first shot and hits a bullseye with his other two shots.
The probability of this happening is 
Add these probabilities:
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
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
3: He misses with his first shot and hits a bullseye with his other two shots.
The probability of this happening is
Add these probabilities:
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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?
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?
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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
.
The probability of hitting the bullseye at least once in three tries is the complement of this, or 
.
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
The probability of hitting the bullseye at least once in three tries is the complement of this, or
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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?
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?
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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
This is the expected value of one ticket.
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
This is the expected value of one ticket.
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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.
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.
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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:
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:
To find our final answer, we need to multiply these two together.
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:
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:
To find our final answer, we need to multiply these two together.
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If you flip a quarter three times, what is the probability of it landing on heads all three times?
If you flip a quarter three times, what is the probability of it landing on heads all three times?
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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:
So there is a 1 in 8 chance that a coin will land on heads three times in a row.
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:
So there is a 1 in 8 chance that a coin will land on heads three times in a row.
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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?
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?
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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:
Now we divide the number of blue socks by the total number of socks to find the probability of picking a blue sock:
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:
Now we divide the number of blue socks by the total number of socks to find the probability of picking a blue sock:
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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?
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?
Tap to reveal answer
We can calculate the two individual probabilities first.
Prob(die shows even) =\frac{3}{6}$=\frac{1}{2}$ (2, 4, and 6 out of 1, 2, 3, 4, 5, and 6)
Prob(tail) =\frac{1}{2}$ (tail out of head and tail)
Then, Prob(even AND tail) is P(even)times P(tail)=\frac{1}{2}$times $\frac{1}{2}$.
We can calculate the two individual probabilities first.
Prob(die shows even) =\frac{3}{6}$=\frac{1}{2}$ (2, 4, and 6 out of 1, 2, 3, 4, 5, and 6)
Prob(tail) =\frac{1}{2}$ (tail out of head and tail)
Then, Prob(even AND tail) is P(even)times P(tail)=\frac{1}{2}$times $\frac{1}{2}$.
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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?
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?
Tap to reveal answer
The easiest way to solve this is by using the complement. Only one of the five plates is NOT blue or green. So $\frac{1}{5}$ of the plates are NOT blue or green. Therefore 1-$\frac{1}{5}$=\frac{4}{5}$ of the plates are blue or green.
The easiest way to solve this is by using the complement. Only one of the five plates is NOT blue or green. So $\frac{1}{5}$ of the plates are NOT blue or green. Therefore 1-$\frac{1}{5}$=\frac{4}{5}$ of the plates are blue or green.
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Two dice are rolled. What is the probability that the sum of both dice is greater than 8?
Two dice are rolled. What is the probability that the sum of both dice is greater than 8?
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There are 36 possible outcomes (). 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).
$\frac{10}{36}$ = $\frac{5}{18}$
There are 36 possible outcomes (
$\frac{10}{36}$ = $\frac{5}{18}$
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