Discrete Probability - GMAT Quantitative

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Question

Fifty marbles are put into a big box: ten red, ten yellow, ten green, and twenty blue. Which of the following actions would change the probability that a random draw would result in a blue marble?

Answer

The ratio of blue marbles to non-blue marbles is . To maintain the same probability of drawing a blue marble, this ratio must remain the same after the action.

We will look at all five actions.

If three yellow marbles and two blue marbles are added, the ratio is .

If fifteen green marbles and ten blue marbles are added, the ratio is .

If two yellow marbles, four green marbles, and four blue marbles are removed, the ratio is .

If half of the marbles of each color are removed - that is, ten blue marbles and five of each of the other three colors - the ratio is .

If three marbles of each color are removed, however, the ratio is $17: 21\neq 2:3$ . This is the correct choice.

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Question

One hundred marbles - red, yellow, blue, or green - are placed in a box. There are an equal number of red and blue marbles, three times as many yellow marbles as green marbles, and five fewer green marbles than red marbles. What is the probability that a randomly drawn marble is NOT blue?

Answer

If there are blue marbles, there are also red marbles, green marbles, and $3 \left (x-5 \right )$ yellow marbles. There are 100 marbles total, so we can solve for in the equation:

20 of the 100 marbles are blue, so the probability that the marble is not blue is:

$\frac{100-20}{100} = \frac{80}{100} = \frac{4}{5}$

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Question

Two eight-sided dice from a role-playing game are thrown. Each die is fair and marked with the numbers 1 through 8. What is the probability that the sum of the dice will be a prime number?

Answer

There are $8 \cdot 8= 64$ possible outcomes. The sum can be between 2 and 16 inclusive; we count the number of rolls that result in the sum being any of the possible prime numbers 2,3,5,7,11,13:

23 out of 64 rolls result in prime sums, so the probability is $\frac{23}{64}$ .

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Question

Two eight-sided dice from a role-playing game are thrown. Each die is fair and marked with the numbers 1 through 8. What is the probability that the sum of the dice will be a multiple of 3?

Answer

There are $8 \cdot 8= 64$ possible outcomes. The outcome can be between 2 and 16 inclusive; we count the number of rolls that result in any of the possible multiples of 3 - 3,6,9,12,15:

22 out of 64 outcomes result in a multiple of 3, so the probability is $\frac{22}{64} = \frac{11}{32}$

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Question

What is the probability of drawing a red king in a standard card deck?

Answer

There are red kings in a standard deck of cards, therefore the probability of drawing red kings is $\frac{2}{52}$ which simplifies to $\frac{1}{26}$ .

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Question

We throw two dice. What is the probability that the sum of the two sides showing up is equal to ?

Answer

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 $6\cdot6$ possible outcomes and the final answer is $\frac{5}{36}$ .

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Question

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?

Answer

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 $\frac{3}{51}$ .

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.

$1\cdot \frac{3}{51}$ is the final answer

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Question

The upper portion of the above spinner has radius twice that of the lower portion.

If the above spinner is spun, what are the odds against the arrow stopping in the purple region?

Answer

The radii of the portions of the spinner - and the areas of the sectors - are actually irrelevant to the problem; it is the measures of their central angles that count.

The purple sector is one-sixth of the larger semicircle, so it is one-twelfth of a circle. This means that the probability of a spinner stopping inside that sector is $\frac{1}{12}$ , and the odds against this are

$\frac{\frac{11}{12} }{\frac{1}{12}} = \frac{11}{1}$ - that is, 11 to 1.

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Question

Race cars for a particular race are numbered sequentially from 12 to 115. What is the probability that a car selected at random will have a tens digit of 1?

Answer

There are 104 integers from 12 to 115 inclusive. There are 8 integers from 12 to 19 and 6 integers from 110 to 115 for a total of 14 integers with a tens digit of 1. The probability of selecting a car with a tens digit of 1 is $\frac{14}{104}$ = $\frac{7}{52}$ .

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Question

Three friends play marbles each week. When they combine their marbles, they have 100 in total. 45 of the marbles are new and the rest are old. 30 are red, 20 are green, 25 are yellow, and the rest are white. What is the probability that a randomly chosen marble is new OR yellow?

Answer

Prob(new OR yellow) = P(new) + P(yellow) - P(new AND yellow)

Prob(new) = $\dpi{100} \small \frac{45}{100}$

Prob(yellow) = $\dpi{100} \small \frac{25}{100}$

Prob(new AND yellow) = $\dpi{100} \small \frac{45}{100}\times \frac{25}{100}$

so P(new OR yellow) = $\dpi{100} \small \frac{45}{100}+\frac{25}{100}-\frac{45}{100}\times \frac{25}{100}$

$\dpi{100} \small =\frac{70}{100}-\frac{9}{80}$

$\dpi{100} \small =\frac{7}{10}-\frac{9}{80}$

$\dpi{100} \small =\frac{56}{80}-\frac{9}{80}=\frac{47}{80}$

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Question

Flight A is on time for 93% of flights. Flight B is on time for 89% of flights. Flight A and B are both on time 87% of the time. What is the probabiity that at least one flight is on time?

Answer

P(Flight A is on time) = 0.93

P(Flight B is on time) = 0.89

P(Flights A and B are on time) = 0.87

Then P(A OR B) = P(A) + P(B) - P(A AND B) = 0.93 + 0.89 – 0.87 = 0.95

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Question

At a school fair, there are 25 water balloons. 10 are yellow, 8 are red, and 7 are green. You try to pop the balloons. Given that you first pop a yellow balloon, what is the probability that the next balloon you hit is also yellow?

Answer

At the start, there are 25 balloons and 10 of them are yellow. You hit a yellow balloon. Now there are 9 yellow balloons left out of 24 total balloons, so the probability of hitting a yellow next is

$\frac{9}{24}=\frac{3}{8}$ .

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Question

A bag has 7 blue balls and 3 red balls. 2 balls are to be drawn successively and without replacement. What is the probability that the first ball is red and the second ball is blue?

Answer

We first have 7 blue and 3 red out of 10, so P(1st ball is red) = $\frac{3}{10}$ . Now, we have pulled a red ball out of the bag, leaving us with 7 blue and 2 red out of 9 total balls. Then P(2nd ball is blue) = $\frac{7}{9}$ . Put this together, P(1st red AND 2nd blue) = $\frac{3}{10}\cdot \frac{7}{9}=\frac{7}{30}$ .

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Question

4 cards are to be dealt successively and without replacement from an ordinary deck of 52 cards. What is the probability of receiving, in order, a spade, a heart, a diamond, and a club?

Answer

There are 4 suits, and each suit has 13 cards, so P(spade) = $\frac{13}{52}$ .

Now there are 51 cards left: 12 spades, 13 hearts, 13 diamonds, and 13 clubs, so now P(heart) = $\frac{13}{51}$ .

Once again, there are now 50 cards: 12 spades, 12 hearts, 13 diamonds, and 13 clubs, so now P(diamond) = $\frac{13}{50}$ .

Before our last draw we have 49 cards left: 12 spades, 12 hearts, 12 diamonds, and 13 clubs, so P(club) = $\frac{13}{49}$ .

Putting these four probabilities together gives us our answer:

P(spade AND heart AND diamond AND club) = $\frac{13}{52 }\cdot \frac{13}{51}\cdot \frac{13}{50}\cdot \frac{13}{49}$

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Question

A red die and a white die are rolled. What is the probability of getting a 4 on the red die AND an odd sum of numbers on the two dice?

Answer

Let's first look at the two probabilities separately.

There are 36 possible combinations of the two dice: (1, 1), (1, 2), (1, 3),...,(1, 6); (2, 1), (2, 2), (2, 3),...; ....; (6, 1), ..., (6, 6). Getting a 4 on the red die can happen 6 different ways: (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), and (4, 6). So P(4 on red die)

$=\frac{6}{36}=\frac{1}{6}$ .

Now, getting an odd sum on the dice can happen 18 different ways: (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (3, 6), (4, 1), (4, 3), (4, 5), (5, 2), (5, 4), (5, 6), (6, 1), (6, 3), and (6, 5). So P(odd sum) =

$\frac{18}{36}=\frac{1}{2}$ .

Putting them together, P(4 on red AND odd sum) =

$\frac{1}{6}\cdot \frac{1}{2}=\frac{1}{12}$ .

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Question

A box contains 12 balls, of which 5 are black, 4 are red, and 3 are white. If 2 balls are randomly selected from the box, one at a time without being replaced, what is the probability that the first ball selected will be red and the second ball selected will be white?

Answer

The probability that the first ball selected will be red is $\frac{4}{12}=\frac{1}{3}$ .

The probability that the second ball selected will be white is $\frac{3}{11}$ .

We can solve by the multiplication principle since the two events happen together.

$\left ( \frac{1}{3} \right )\times\left ( \frac{3}{11} \right )=\frac{1}{11}$

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Question

What is the probability of sequentially drawing 2 diamonds from a deck of regular playing cards when the first card is not replaced?

Answer

The probability of drawing a diamond first is $\frac{13}{52}=\frac{1}{4}$

The probabilty of drawing a diamond second when the first card is not replaced is $\frac{12}{51}$ .

To determine the probability of 2 INDEPENDENT events, we multiply them.

$P=\frac{1}{4}\cdot \frac{12}{51}=\frac{12}{204}=\frac{1}{17}$

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Question

Two fair dice are tossed. What is the probability that the dice will add up to a prime number?

Answer

The prime numbers that can be rolled are 2, 3, 5, 7, and 11. The rolls that result in one of these numbers, out of a possible $6 \cdot 6 = 36$ rolls, are:

Therefore, there are 15 rolls out of 36 that result in a prime number, making the probability of a prime result $\frac{15}{36} = \frac{5}{12}$

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Question

Two fair six-sided dice are rolled. What is the probability that the sum is either a perfect square or a perfect cube?

Answer

There are 36 possible rolls. The only perfect squares that can be rolled are 4 and 9; the only perfect cube that can be rolled is 8. So the roll must be one of the following:

This adds up to 12 rolls out of 36, for a probability of $\frac{12}{36} = \frac{1}{3}$

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Question

Two coins are loaded so that each comes up heads 60% of the time when tossed. If the coins are both tossed at the same time, what is the probability that the result will be one head and one tail?

Answer

For each coin, the probability of a head is 0.6, and the probability of a tail is 0.4.

The probability of a head coming up on the first coin and a tail coming up on the second is $0.6 \cdot 0.4 = 0.24$ ; the same holds for the reverse case. Therefore, the probability of one head and one tail is

$\ 0.24 + 0.24 = 0.48$

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