Calculating discrete probability - GMAT Quantitative

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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 ?

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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?

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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?

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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?

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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?

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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?

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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?

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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?

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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?

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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?

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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?

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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?

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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?

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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?

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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?

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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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Question

A die is loaded so that it is equally likely to come up 1, 2, 3, 4, or 5, but twice as likely to come up 6 as it is likely to come up 5. If it is rolled twice, what is the probability that both rolls are even?

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Answer

Let be the probability that the die will come up 1. Then is also the common probability for each of the rolls of 2, 3, 4, or 5, and is the probability of the die coming up 6. To determine the value of , add these six probabilities and set the sum to 1, then solve:

$5 \cdot P + 2P = 1$

$P = \frac{1}{7}$

So 1, 2, 3, 4, and 5 each will come up with probability $\frac{1}{7}$ and 6 will cme up with probability $\frac{2}{7}$ .

An even number will come up with probability $\frac{1}{7} + \frac{1}{7} + \frac{2}{7} = \frac{4}{7}$ . Two even rolls will happen with probability $\frac{4}{7 } \cdot \frac{4}{7 } = \frac{16}{49}$ .

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Question

Which of the following can be done to a standard deck of fifty-two playing cards without affecting the probability that a randomly chosen card will be a club?

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Answer

The probability of drawing a club from a standard deck of cards is $\frac{1}{4}$ , since one-fourth of the cards - thirteen out of fifty-two - are clubs.

If the two red kings are removed, thirteen out of the remaining fifty cards are clubs, and the probability is $\frac{13}{50} \neq \frac{ 1}{4}$ .

If the thirteen hearts are removed, thirteen out of the remaining thirty-nine cards are clubs, and the probability is $\frac{13}{39}= \frac{ 1}{3} \neq \frac{ 1}{4}$ .

If the ace of spades is removed, thirteen out of the remaining fifty-one cards are clubs, and the probability is $\frac{13}{51} \neq \frac{ 1}{4}$ .

If the joker is added, thirteen out of the fifty-three cards are clubs, and the probability is $\frac{13}{53} \neq \frac{ 1}{4}$ .

If the four fours are removed, this leaves twelve clubs out of forty-eight cards, so the probability is $\frac{12}{48} = \frac{1}{4}$ . This is the corect choice.

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Question

One hundred marbles - each one red, yellow, blue, or green - are placed in a box. Forty of the marbles are green, there are twice as many blue marbles as there are red ones, and there are three times as many yellow marbles as red ones. What is the probability that a randomly drawn marble will be yellow?

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Answer

Let be the number of red marbles. Then there are blue marbles and yellow ones. Since there are 40 green marbles, and 100 marbles total, we can write this equation, simplifying and solving for :

$6x \div 6 = 60\div 6$

Therefore, there are $3x = 3 \cdot10 = 30$ yellow marbles, and the probability of drawing a yellow marble is $\frac{30}{100} = \frac{3}{10}$

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Question

Four coins are tossed at the same time. Three of them are fair; the fourth is loaded so that it comes up heads with probability . In terms of , what is the probability that the outcome will be three heads and one tail?

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Answer

Call the fair coins, which come up heads or tails with probability $\frac{1}{2}$ , Coins 1, 2, and 3; call the loaded coin, which comes up heads with probability and tails with probability , Coin 4. We are looking for the probability of one of the following four outcomes:

The probability of heads on Coins 1, 2, and 3 and tails on Coin 4:

$\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot (1 -N) = \frac{1}{8} \cdot (1 -N) = \frac{1}{8} -\frac{1}{8} N$

The probability of tails on Coin 1 and heads on Coins 2, 3, and 4:

$\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}\cdot N= \frac{1}{8} N$

The probability of tails on Coin 2 and heads on Coins 1, 3, and 4:

$\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}\cdot N= \frac{1}{8} N$

The probability of tails on Coin 3 and heads on Coins 1, 2, and 4:

$\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}\cdot N= \frac{1}{8} N$

Add these:

$\frac{1}{8} N +\frac{1}{8} N + \frac{1}{8} N + \frac{1}{8} -\frac{1}{8} N$

$= \frac{1}{8} N +\frac{1}{8} N + \frac{1}{8}$

$= \frac{1}{4} N + \frac{1}{8}$

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Question

Three coins are tossed at the same time. One of them is fair; two are loaded so that each comes up heads with probability . In terms of , what is the probability that the outcome will be three heads?

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Answer

The probability that all three coins will come up heads is the product of the individual probabilities:

$\frac{1}{2}\cdot N \cdot N = \frac{1}{2}N ^{2} = \frac{N ^{2} }{2}$

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