GRE Quantitative Reasoning - How to find the probability of an outcome

Presented with a deck of fifty-two cards (no jokers), what is the probability of drawing either a face card or a spade?

$\frac{11}{26}$

$\frac{25}{52}$

$\frac{156}{2704}$

$\frac{3}{52}$

$\frac{9}{26}$

Explanation

A face card constitutes a Jack, Queen, or King, and there are twelve in a deck, so the probability of drawing a face card is $P(F)=\frac{12}{52}$ .

There are thirteen spades in the deck, so the probability of drawing a spade is $P(S)=\frac{13}{52}$ .

Keep in mind that there are also three cards that fit into both categories: the Jack, Queen, and King of Spades; the probability of drawing one is $P(F\bigcap S)=\frac{3}{52}$

Thus the probability of drawing a face card or a spade is:

$P(F)+P(S)-P(F\bigcap S)=\frac{12+13-3}{52}=\frac{11}{26}$

The probability that event A occurs is 0.22. The probability of event B occuring is 0.35. If the probability of event C occuring is 0.33, and the probability of event A or event B occuring is 0.54, then the probability of event C occuring is how many times the probability of event A and B occuring?

Explanation

Keywords in this question are or and and. Here, the probability of event A or B is p(A) + p(B) if they are mutually exclusive. If they are not, however, the probability of A or B is p(A) + p(B) - p(A and B).

Mutually exclusive: $p(A\ or\ B)=0.22+0.35=0.57$

Not mutually exclusive: $p(A\ or\ B)=0.22+0.35-p(A\ and\ B)$

Given that p(A) + p(B) = 0.57, and we are told that p(A or B) = 0.54, it implies that p(A and B) = 0.03 (and that A and B are not mutually exclusive).

$p(A\ or\ B)=0.54=0.22+0.35-p(A\ and\ B)=0.57-p(A\ and\ B)$

$p(A\ and\ B)=0.03$

Now we can solve to ratio of p(A and B) to p(C).

$\frac{p(C))}{p(A\ and\ B)}=\frac{0.33}{0.03}=11$

A coin is flipped four times. What is the probability of getting heads at least three times?

$\frac{5}{16}$

$\frac{1}{16}$

$\frac{1}{4}$

$\frac{11}{16}$

$\frac{3}{4}$

Explanation

Since this problem deals with a probability with two potential outcomes, it is a binomial distribution, and so the probability of an event is given as:

$P(n,k)=\frac{n!}{(n-k)!k!}p^k(1-p)^{n-k}$

Where is the number of events, is the number of "successes" (in this case, a "heads" outcome), and is the probability of success (in this case, fifty percent).

Per the question, we're looking for the probability of at least three heads; three head flips or four head flips would satisfy this:

$P(4,3)=\frac{4!}{(4-3)!3!}(\frac{1}{2})^{3}(\frac{1}{2})^{4-3}=4(\frac{1}{2})^{4}=\frac{1}{4}$

$P(4,4)=\frac{4!}{(4-4)!4!}(\frac{1}{2})^{4}(\frac{1}{2})^{4-4}=(\frac{1}{2})^{4}=\frac{1}{16}$

Thus the probability of three or more flips is:

$\frac{1}{4}+\frac{1}{16}=\frac{5}{16}$

A coin is flipped seven times. What is the probability of getting heads six or fewer times?

$\frac{127}{128}$

$\frac{255}{256}$

$\frac{1}{128}$

$\frac{1}{7}$

$\frac{6}{7}$

Explanation

Since this problem deals with a probability with two potential outcomes, it is a binomial distribution, and so the probability of an event is given as:

$P(n,k)=\frac{n!}{(n-k)!k!}p^k(1-p)^{n-k}$

Where is the number of events, is the number of "successes" (in this case, a "heads" outcome), and is the probability of success (in this case, fifty percent).

One approach is to calculate the probability of flipping no heads, one head, two heads, etc., all the way to six heads, and adding those probabilities together, but that would be time consuming. Rather, calculate the probability of flipping seven heads. The complement to that would then be the sum of all other flip probabilities, which is what the problem calls for:

$P(7,7)=\frac{7!}{(7-7)!7!}(\frac{1}{2})^7(1-\frac{1}{2})^{7-7}=(\frac{1}{2})^7=\frac{1}{128}$

Therefore, the probability of six or fewer heads is:

$1-\frac{1}{128}=\frac{127}{128}$

A random variable is normally distributed with a mean of 500 and a standard deviation of 20.

---

Probability of the event that

---

$\frac{1}{10}$

Quantity A is greater

Quantity B is greater

The two quantities are equal

The relationship cannot be determined

Explanation

In a normally distributed curve, a standard deviation of 1 contains ~68% of all values within its range, and a standard deviation of 2 contains ~95% of all values within its range. Since the mean value is 500 for this situation, two standard deviations would occur at values 460 to 540 (i.e. ~95% of all values are within this range). This would mean that the value of 450 would fall outside this range, i.e. a 100%–95% = 5% probability. Since 5% expressed as a fraction is 1/20, Quantity B is far greater than the probability of landing at 450.

There are 31 students in the library.

Quantity A: The probability that all students have their birthdays in January. (There are 31 days in January).

Quantity B: The probability that all students have their birthdays on a Saturday this year. (Assume there are 52 Saturdays in a year).

Quantity A is greater.

Quantity B is greater.

The two quantities are equal.

The relationship cannot be determined from the information given.

Explanation

As the options suggest, this problem requires us to know how probabilities work. The probability of a single event happening is equal to the number of ways it can happen divided by the total number of outcomes. Because this problem asks about 31 students, we have to raise our probability to the 31st power because it's the probability of 31 different events happening.

Now we can figure out the probability of one January birthday (31 days in January/365 days in a year) and the probability of all Saturday birthdays (52 Saturdays in a year/365 days in a year). So thus the probabilities that ALL of the students will have either type of birthday is:

(31/365)31 and

(52/365)31

Since the numbers are raised to the same power, we can simply look at the base to determine which is larger and which is smaller. Since the probability of having a birthday on Saturday (roughly 1/7) is greater than having a birthday in January (roughly 1/12), then Quantity B is greater.

The probability that events A and/or B will occur is 0.88.

Quantity A: The probability that event A will occur.

Quantity B: 0.44.

Quantity A is greater.

Quantity B is greater.

The two quantities are equal.

The relationship cannot be determined from the information given.

Explanation

The only probabilites that we know from this is that P(only A) + P(only B) + P (A and B) = 0.88, and that P(neither) = 0.12. We cannot calculate the probability of P(A) unless we know two of the probabilites that add up to 0.88.

A high school has 200 students. 120 are male, 50 are upper division students, and 40 are upper division male students. What is the probability of choosing a lower division female student, given the student is female?

7/8

7/15

7/20

1/4

2/5

Explanation

There are 200 students in total, and 120 of them are male, so 80 must be female. We also know that there are 50 upper division students, and 40 of them are male, so 10 must be female. If 10 of 80 females are upper division, the other 70 females have to be lower division students, so the probability of choosing a lower division student, given the student is female, is 70/80 = 7/8.

A fair six-sided die is thrown.

Quantity A: The probability that the die will show either a "6" or an odd number.

Quantity B: $\dpi{100} \small \frac{2}{3}$

The two quantities are equal.

Quantity A is greater.

Quantity B is greater.

The relationship cannot be determined from the information given.

Explanation

To obtain probability, you must divide the number of possible desired outcomes by the number of total possible outcomes. The number of desired outcomes is 4 (3 odd numbers plus 1 even number). The number of total outcomes is 6 (number of sides on the die). So $\dpi{100} \small \frac{4}{6}$ reduces to $\dpi{100} \small \frac{2}{3}$ , making the 2 quantities equal.

At Jill's school fair, there is a game with 25 balloons hung on a dart board. 10 are blue, 8 are red, and 7 are green. Jill throws a dart and pops a blue balloon. What is the probability that the next balloon she hits will also be blue?

2/5

3/8

3/5

9/25

5/12

Explanation

Since one blue balloon has already been popped, there are now 9 blue balloons left, and 24 balloons left overall. Therefore the probability that the next balloon Jill hits is also blue is 9/24 = 3/8.

How to find the probability of an outcome

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