All GMAT Math Resources
Example Questions
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 .
Example Question #41 : Calculating Discrete Probability
We throw two dice. What is the probability that the sum of the two sides showing up is equal to ?
In this problem, the best way to calculate the answer is to count the possible outcomes. We can see that the sum can be obtained with the following sums: 2+6,3+5, 4+4, 5+3, 6+2 .
We have a total of 5 outcomes. As previously discussed, some sums are counted twice because in probabilty we study events as if they were successive events even when they are not. For this same reason, the 4+4 is counted only once, because a dice yielding a 4 followed by a dice yielding a 4 is the same as a dice yielding a 4 followed by a dice yielding a 4.
There are a total of possible outcomes and the final answer is .
Example Question #42 : Calculating Discrete Probability
Let's try to solve more complicated problems using the same logic as in the dice problems.
We have a standard -card deck. We draw two cards without replacement. What is the probabilty of drawing a pair?
Here we are asked for the probability of drawing any pair. Therefore, the first card can be any card and we can assign this event a probability of one.
Furthermore, the second card drawn must be the same card to form a pair. Therefore, its probability is .
We use 51 because we already have drawn a card. You will notice that to form a pair there are 3 other cards in the deck that can be drawn, for example if an ace is drawn first, there must be three other aces in the deck.
is the final answer
Example Question #291 : Arithmetic
We have a standard -card deck. We draw two cards without replacement. What is the probabilty of drawing a pair of aces?
In order to draw a pair of ace, the first card drawn must be an ace, which has a probability of being drawn of .
Since there are 4 aces in the deck. The second ace has a probability of being drawn of , since there are only three aces left in the deck.
Therefore to obtain the probability of these two successive events, we have to multiply each probability.
The final answer is given by , or
Example Question #292 : Arithmetic
We have a regular -card deck. We draw cards, without replacement, what is the probabilty that they are all of the same suit (either all spades, all hearts, all clubs or all diamonds)?
First of all, the first card can be any card, so the probabilty of drawing any card is 1, since it is necessarily a success. The second card must belong to the same suit (don't be confused by the fact that the cards can be red or black, whenever color is mentioned it is said to reference the suits), in total there are 13 cards by colors. Since we have already drawn one card, there are only 12 left out of a total of 51 cards, after we have drawn two card of the same suit, there will be only 11 cards of this suit left out of 50.
The probability is given by the following calculation :
Example Question #293 : Arithmetic
A woman can shoot a rifle in the desert and successfully hit the target with a probabilty of . If she shoots times in a row what is the probability that she hits the target only once?
The answer can be obtained by firstly calculating the probabilty of her missing twice and hitting the target once, in any order.
This probability is obtained by calculating .
is the probability of her failing, which is calculated by substracting .
Example Question #294 : Arithmetic
A woman shoots her rifle in the desert, she hits the target times out of . What is the probabilty that she hits the target at least once if she shoots times in a row?
The answer can be obtained by finding the opposite of the event 'she hits at least one target'. This event is 'the woman misses all her 4 shots'.
This probabilty is calculated by raising to the power of 4, since she misses all her 4 shots.
The result is then substracted from 1, 1 being the total of all the possible probabilities for all events in a given problem.
, which is the final answer.