GMAT Quantitative › Calculating discrete probability
Presented with a deck of fifty-two cards (no jokers), what is the probability of drawing either a face card or a spade?
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 .
There are thirteen spades in the deck, so the probability of drawing a spade is .
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
Thus the probability of drawing a face card or a spade is:
A coin is flipped four times. What is the probability of getting heads at least three times?
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:
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:
Thus the probability of three or more flips is:
A coin is flipped seven times. What is the probability of getting heads six or fewer times?
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:
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:
Therefore, the probability of six or fewer heads is:
Some balls are placed in a large box; the balls include one ball marked "A", two balls marked "B", and so forth up to twenty-six balls marked "Z". A ball is drawn at random.
Given a particular letter of the alphabet, does the probability that that ball will be marked with that letter exceed ?
Statement 1: The letter appears in the word "lousy".
Statement 2: The letter appears in the word "skunk".
STATEMENT 1 ALONE provides sufficient information to answer the question, but STATEMENT 2 ALONE does NOT provide sufficient information to answer the question.
STATEMENT 2 ALONE provides sufficient information to answer the question, but STATEMENT 1 ALONE does NOT provide sufficient information to answer the question.
EITHER STATEMENT ALONE provides sufficient information to answer the question.
BOTH STATEMENTS TOGETHER provide sufficient information to answer the question, but NEITHER STATEMENT ALONE provides sufficient information to answer the question.
BOTH STATEMENTS TOGETHER do NOT provide sufficient information to answer the question.
The total number of balls in the box will be
.
Since
,
it follows that the number of balls is
.
The number of balls with a given letter of the alphabet is equal to the number of its position in the alphabet; the probability of a ball with that letter being drawn is that number divided by the total number of balls, 351. Therefore, for this probability to exceed
, we must have the relation
.
Therefore, .
The 11th letter of the alphabet is "K", so in order to answer this question, it suffices to know whether the letter comes after "K" in the alphabet.
From Statement 1 alone, the question can be answered, since all of the letters in the word "lousy" appear after "K" in the alphabet. From Statement 2 alone, however, the question cannot be answered, since the letter "K" itself appears in the word "skunk".
Some balls are placed in a large box, which include ball marked "A",
balls marked "B",
balls marked "C", and so on up to
balls marked "Z".
A ball is drawn at random. Give the probability that the ball will be marked "O".
The total number of balls in the box will be
.
Since
,
it follows that the number of balls is
.
Since "O" is the fifteenth letter of the alphabet, there are balls out of
marked with "O". Therefore, the probability that a randomly-drawn ball will be one of these is
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
.
The upper semicircle of the above target has radius twice that of the lower semicircle.
A blindfolded woman throws a dart at random at the above target. Let be the probability that the dart lands inside a green region; let
and
be similarly defined for a red or yellow region. Which of the following is a true statement?
(Assume that skill is not a factor, and that the dart hits the target.)
Since the areas of the sectors are what are important here, we need only rank the total areas of the different colors, which we will call .
For the sake of simplicity, we will assume the smaller semicircle of the target has radius ; consequently, the larger semicircle has radius
.
The larger semicircle has total area
The smaller semicircle has total area
Each of the six sectors of the larger semicircle has area of its area, or
Each of the two sectors of the smaller semicircle has area of its area, or
The total of the areas of the green sectors - three sectors of the larger semicircle - is
The total of the areas of the red sectors - two sectors of the larger semicircle and one of the smaller - is
The total of the areas of the yellow sectors - one sector of the larger semicircle and one of the smaller - is
; it follows that
.
Some balls are placed in a large box, which include one ball marked , two balls marked
, and continuing up to
balls marked
.
What is the probability that a ball randomly drawn from this box will be marked with a prime number?
The total number of balls in the box will be
.
Since,
,
it follows that the number of balls is
.
The prime numbers in the range of are
. The number of balls with each number corresponds to the number itself, so the number of balls with prime numbers will be
.
The probability of drawing one of them is .
Some balls are placed in a large box, which include one ball marked "A", two balls marked "B", three balls marked "C", and so forth up to twenty-six balls marked "Z".
A ball is drawn at random. Give the odds against the ball being marked "I".
The total number of balls in the box will be
.
Since
,
it follows that the number of balls is
.
Since "I" is the ninth letter of the alphabet, there are nine balls marked "I", and other balls. Therefore, the odds against drawing a ball marked "I" are 342 to 9, or
The correct choice is 38 to 1.
A coin is tossed times, what is the probabilty of having a head showing up on the third toss?
This trick question is in fact very simple. Since the throws are independent (in other words, what happened before an event has no influence on the future outcomes), then the probabilty is , indeed there are no reasons for any of the head or tail to have more chance of showing up at any point.