All GMAT Math Resources
Example Questions
Example Question #31 : Discrete Probability
Scientists tried out a new way of testing people for a certain disease. Out of patients tested, tested positive when they did not have the disease. On the other hand, tested negative when they in fact had the disease. What is the probability that a randomly selected patient was accurately tested (i.e positive if he or she actually had the disease and negative if he or she did not)?
Out of the patients, wrongly tested positive and wrongly tested negative. There is a total of patients whose tests were not accurate.
The probability that a randomly selected patient was accurately tested is:
Example Question #1831 : Problem Solving Questions
John alters a fair coin so that the probability of it coming up heads is 5% greater than that of it coming up tails. By how much has he increased the probability of it coming up heads?
The probablilty of a fair coin coming up heads is .
If the coin is altered as in the problem, then the odds of the coin coming up heads as opposed to tails are 105 to 100; restated, the probability of it coming up heads is
John has increased the probability of the coin coming up heads by
Example Question #32 : Calculating Discrete Probability
Marianna inserts the joker into a standard deck of 52 cards. By how much has she decreased the probability that a card drawn at random from the deck will be an ace?
The probability that a card, randomly drawn from a standard deck of 52 cards without the joker, is an ace (or any other given rank) is ; if the joker is added, the probability decreases to . The decrease in probability is
Example Question #33 : Calculating Discrete Probability
Pilar inserts the joker into a standard deck of 52 cards. By how much has she decreased the probability that a card drawn at random from the deck will be a diamond?
None of the other responses gives the correct answer.
The probability that a card, randomly drawn from a standard deck of 52 cards without the joker, is a diamond (or any other given suit) is ; if the joker is added, the probability decreases to . The decrease in probability is
.
Example Question #34 : Calculating Discrete Probability
Judy alters a fair six-sided die so that a roll of 1, 2, 3, 4, or 5 are still equally likely, but a roll of 6 is 5% more likely than any one of the other rolls.
By how much has she increased the probability of rolling a 6?
The probability of rolling a 6 with a fair die is .
Since a roll of 6 is 5% more likely than any one of the other rolls, and the other rolls are equally probable, then, theoretically, there should be 105 rolls of 6 for every 100 rolls of 5, 100 rolls of 4, etc. - that is, 105 out of 605, rolls should theoretically be a 6. This makes the probability of rolling a 6 on the altered die
.
Therefore, Judy's alteration has increased the probability of rolling a 6 by
Example Question #1836 : Gmat Quantitative Reasoning
Carla alters a fair six-sided die so that a roll of 1, 2, 3, 4, or 5 are still equally likely, but a roll of 6 is 4% more likely than any one of the other five outcomes.
By how much has she decreased the probability of rolling a 5?
The probability of rolling a 5 with a fair die is .
Since a roll of 6 on the altered die is 4% more likely than any one of the other rolls, and the other rolls are equally probable, then, theoretically, there should be 104 rolls of 6 for every 100 rolls of 5, 100 rolls of 4, etc. Consequently, 100 out of 604 rolls should theoretically result in a 5. This makes the probability of rolling a 5 on the altered die
.
Therefore, Pilar's alteration has decreased the probability of rolling a 5 by
Example Question #1837 : Gmat Quantitative Reasoning
Mick alters a fair coin so that the probability of it coming up heads is 4% greater than that of it coming up tails. By how much has he decreased the probability of it coming up tails?
The probablilty of a fair coin coming up tails is .
If the coin is altered as in the problem, then the odds of the coin coming up tails as opposed to heads are 100 to 104; restated, the probability of it coming up tails is
Mick has decreased the probability of the coin coming up tails by
Example Question #41 : Discrete Probability
This problem set is designed to have a better understanding of probabilties
We throw two regular six sided dice. What is the probabilty of having a pair?
The correct answer is , since there are 2 dice and we are looking for the probabilty than any pair shows up, the first die can then be any number, therefore we assign it a probability of one and the second die must be the same number as the first, and therefore has a probabilty of . Multiplying these two probabilities, we obtain .
Example Question #1831 : Problem Solving Questions
We throw two regular six sided dice. What is the probabilty of having a pair of one?
In this problem, we are asked for the probabilty that a specific pair shows up. The probability that the number one shows up on a single die is . Here, we want this event 'the 1 shows up' to occur twice. The result is given by calculating which is .
Remember, the probability of two succesful events is calculated by multiplying their individual probabilities.
Example Question #283 : Arithmetic
We throw two six-sided dice. What is the probabilty that the sum of the values showing up is equal to ?
The best way to solve this problem is to count the possible outcomes for this event. Indeed, a result of five is given by different sums; 1+4, 2+3, 3+2, 1+4. The fact that we count twice each of these sum may seem counterintuitive but it stems from the fact that in probabilty we consider events to occur successively, whereas in reality the dice have been thrown at the same time. But if they were thrown one after the other, we would note that obtaining a 2 after 3 is different that obtaining a 3 after a 2.
We then have 4 possible outcomes for the success of the event 'the sum is 5'. The total number of possible outcomes is given by the multiplication of all the possible values on each die or , which is equal to 36.
The final answer is obtained by dividing the number of possible outcomes by the total number of outcomes, in other words by calculating .