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Example Questions
Example Question #122 : Probability
Forty balls are placed in a large box, each marked with a whole number from 1 to 40. The balls marked with an even number or a prime number are all blue; the other balls are red.
A ball is drawn and replaced; the process is repeated. Give the probability that a blue ball is drawn both times.
There are twenty even whole numbers from 1 to 40 - 2, 4, 6, up to 40. Also, there are eleven odd primes - 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37 (there is one even prime, 2, but it has already been counted). Therefore, the box will contain 31 blue balls out of 40, and the probability that the draw of one ball will result in the ball being blue is . This same probability holds in both the first and second draws.
Since the draws are independent events, the probabilities can be multiplied:
,
the correct choice.
Example Question #124 : Probability
Twenty-six balls are placed in a box, each with a different letter of the alphabet. All twenty-six letters are represented.
A ball is drawn, then replaced; the process is repeated two more times. What is the probability that a ball with a vowel is drawn all three times?
(Note: "Y" is to be considered a consonant here.)
None of the other responses gives a correct answer.
5 of the 26 balls are marked with vowels, so the probability that the draw of one ball will result in the ball being marked with a vowel is ; this probability is the same for each of the three draws. Since the draws are independent events, the probabilities can be multiplied:
.
Example Question #121 : Probability
Balls are placed in a large box. Each ball is marked with a whole number from 1 to 20; each even number is represented by two balls, and each odd number is represented by one ball.
A ball is drawn, then replaced; the process is then repeated. Give the probability that an even number will be drawn both times.
There are 10 even numbers and 10 odd numbers in the set of whole numbers from 1 to 20. Therefore, there are 20 even-numbered balls and 10 odd-numbered balls - a total of 30 balls in the box.
The probability that the draw of one ball will result in an even-numbered ball is
This probability is the same for each of the two draws. Since the draws are independent events, the probabilities can be multiplied:
,
the correct response.
Example Question #121 : How To Find The Probability Of An Outcome
Balls are placed in a large box. Each ball is marked with a whole number from 1 to 20; each even number is represented by two balls, and each odd number is represented by one ball.
Two balls are drawn without replacement. Give the probability that both balls will be marked with an odd number.
None of the other responses gives a correct answer.
There are 10 even numbers and 10 odd numbers in the set of whole numbers from 1 to 20. Therefore, there are 20 even-numbered balls and 10 odd-numbered balls - a total of 30 balls in the box.
Assume that the balls are drawn one after the other. The probability that the first ball will be odd is .
Since the ball is not replaced, there are now 29 balls total, 9 of which are marked with odd numbers. The probability that the second ball will be odd is .
Multiply these probabilities to get the probability that both balls will be marked with odd numbers:
.
Example Question #124 : Probability
Twenty-six balls are placed in a box, each with a different letter of the alphabet. All twenty-six letters are represented.
Three balls are drawn without replacement. What is the probability that all three letters on the balls will be consonants?
(Note: "Y" is to be considered a consonant here.)
None of the other responses gives a correct answer.
We can assume that the three balls are drawn one at a time.
When the first ball is drawn, there are 26 balls in the box, 21 of which are marked with consonants. The probability of drawing a ball marked with a consonant is therefore .
Since there is no replacement, when the second ball is drawn, there are 25 balls in the box, 20 of which are marked with consonants. The probability of drawing a ball marked with a consonant is therefore .
Similarly, when the third ball is drawn, there are 24 balls in the box, 19 of which are marked with consonants. The probability of drawing a ball marked with a consonant is therefore .
Multiply the three probabilities together. The probability that all three balls will be marked with consonants is
.
Example Question #131 : Probability
Forty balls are placed in a large box, each marked with a whole number from 1 to 40. The balls marked with an even number or a prime number are all blue; the other balls are red.
Two balls are drawn without replacement. Give the probability that both will be red.
The odd composite numbers - that is, the odd numbers that have more than two factors - are all represented by red balls. The odd composite whole numbers in the set comprise the set - a set of eight elements. The ball marked "1" is also red, since 1 is odd and neither composite nor prime. Therefore, 9 balls of the 40 are red.
We can assume that the two balls are drawn one at a time. The probability of the first ball being red is . Since there are now 39 balls, 8 of which are red, the probability that the second ball will be red is . The probabilities are multiplied:
.
This is the probability that both balls will be red.
Example Question #131 : Outcomes
A popular word game has 100 square tiles. Each is marked with a letter or a blank; the distribution of the letters is seen in the table above. Note that two of the tiles are blanks.
Two letters are drawn at random from the set of tiles. What is the probability that both tiles will be marked with a letter from the first half of the alphabet?
None of the other choices gives the correct response.
Add the numbers of tiles corresponding to each of the letters from "A" to "M":
The probability that one of these tiles will be drawn on the first draw is . On the second draw, there will be 52 such tiles out of 99, making the probability of the same event . The probability of both draws resulting in one of these letters is the product of these individual probabilities:
Example Question #131 : Probability
A popular word game has 100 square tiles. Each is marked with a letter or a blank; the distribution of the letters is seen in the table above. Note that two of the tiles are blanks.
Two letters are drawn at random without replacementfrom the set of tiles. What is the probability that neither tile is marked with a consonant?
(Note: "Y" is considered a consonant for purposes of this problem.)
Out of the 100 tiles, there are 9 tiles marked with "A", 12 with "E", 9 with "I", 8 with "O", and 4 with "U", as well as 2 blank tiles; this is a total of
tiles not marked with a consonant.
The probability that the first draw will result in a vowel or a blank is .
After this draw, there will be 99 tiles, 43 of which are vowels or blanks; the probability that the second draw will be one of these is .
Use the multiplication principle to find the probability that both tiles will be vowels or blanks:
Example Question #134 : Probability
A six-sided die is rolled twice. What is the probability that the first roll will land on an odd number, and the second roll will land on a even number?
A six-sided die has three sides with odd numbers and three sides with even numbers. Thus, the probability of the die landing on an odd number is
which simplifies to
.
The probability of the die landing on an even number is also
which we can simplify to
.
Because the probabilities are independent of each other, we can find the probability of the two events happening in a row by multiplying the fractions.
Therefore, the probability of the first roll landing on an odd number, and the second roll landing on an even number is .
Example Question #111 : Data Analysis
You went to order a cake but all of the cakes are in identical boxes. If there are five chocolate cakes, four carrot cakes, three vanilla cakes, and six red velvet cakes, what is the maximum number of boxes that you would have to open to ensure that you have one of each type of cake?
9
4
12
16
16
There are four different types of cake. In this type of problem we want to guarantee we have one of each, so we need to assumbe we have very bad luck. We start with the red velvet since that is the type with the most cakes. If we open those 6 we are not guaranteed to have different ones. Then say we opened all five chocolate cakes, then all four carrot cakes. We still have only three types of cakes but opened 15 boxes. When we open the next box (16) we will be guaranteed to have one of each.
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