CALCU-He de sortejar 9 problemes per fer entre 20

He de sortejar 9 problemes per fer entre 20 estudiants, de manera que a cada estudiant li pot tocar fer més d'un problema. Quina és la probabilitat que a un estudiant li toqui més d'un problema? Heu de donar el resultat com un nombre real arrodonit a 3 xifres decimals, separant la part entera de la part decimal amb un punt (no amb una coma).

Solution by Steps

step 1

Calculate the total number of ways to distribute 9 problems among 20 students with no restrictions. This is equivalent to putting 9 indistinguishable balls into 20 distinguishable boxes, which can be done in

\binom{9+20-1}{9}

ways

step 2

Calculate the number of ways to distribute 9 problems among 20 students such that each student gets at most one problem. This is equivalent to choosing 9 out of 20 students, which can be done in

\binom{20}{9}

ways

step 3

Calculate the probability that a student gets more than one problem by subtracting the probability of each student getting at most one problem from 1. The probability of each student getting at most one problem is the number of ways to distribute the problems with this restriction divided by the total number of ways to distribute the problems

step 4

Compute the probability using the numbers from steps 1 and 2:

P(\text{more than one problem}) = 1 - \frac{\binom{20}{9}}{\binom{9+20-1}{9}}

step 5

Calculate the numerical value of the probability and round it to three decimal places

Answer

The probability that a student gets more than one problem is approximately 0.849.

Key Concept

Calculating the probability of an event by considering the total number of outcomes and the number of favorable outcomes.

Explanation

The probability that a student gets more than one problem is found by considering all possible distributions of problems and subtracting the probability of the restricted case where each student gets at most one problem.

He de sortejar 5 problemes per fer entre 20 estudiants, de manera que a cada estudiant li pot tocar fer més d'un problema. Quina és la probabilitat que a un estudiant li toqui més d'un problema? Heu de donar el resultat com un nombre real arrodonit a 3 xifres decimals, separant la part entera de la part decimal amb un punt (no amb una coma).

Solution by Steps

step 1

Calculate the total number of ways to assign 5 problems to 20 students with repetition allowed:

N_{\text{total}} = 20^5

step 2

Calculate the number of ways to assign 5 problems to 20 students such that no student gets more than one problem:

N_{\text{no repetition}} = \frac{20!}{(20-5)!}

step 3

Calculate the probability that a student gets more than one problem by subtracting the probability of no student getting more than one problem from 1:

P(\text{more than one}) = 1 - \frac{N_{\text{no repetition}}}{N_{\text{total}}}

step 4

Compute the values:

P(\text{more than one}) = 1 - \frac{20!}{15! \cdot 20^5}

step 5

Simplify the expression and calculate the probability to three decimal places:

P(\text{more than one}) \approx 1 - \frac{20 \cdot 19 \cdot 18 \cdot 17 \cdot 16}{20^5}

Answer

P(\text{more than one}) \approx 1 - \frac{20 \cdot 19 \cdot 18 \cdot 17 \cdot 16}{20^5} \approx 0.893

Key Concept

Calculating probabilities with and without repetition in combinatorial problems

Explanation

The probability that a student gets more than one problem is found by considering all possible distributions of problems and subtracting the probability that each student gets exactly one problem. This involves combinatorial calculations with and without repetition.

What is the total number of possible ways to assign each student to one or more problems out of the 5 given problems?

Solution by Steps

step 1

To determine the total number of ways to assign each student to one or more problems out of the 5 given problems, we consider that each problem can be assigned to a student or not, which is a binary choice (yes or no). Since there are 5 problems, each student has

2^5

ways to be assigned problems, excluding the case where a student is not assigned any problem

step 2

Subtract the one case where a student is not assigned any problem from the total number of ways to assign problems. This is because the question specifies that each student must be assigned to one or more problems.

2^5 - 1 = 32 - 1 = 31

step 3

If there is more than one student, and assuming students are distinguishable, the total number of ways to assign problems to all students would be the number of ways to assign problems to one student raised to the power of the number of students. If

n

is the number of students, the total number of ways would be

31^n

Answer

Without the exact number of students provided, the total number of possible ways to assign each student to one or more problems out of the 5 given problems is

31^n

, where

n

is the number of students.

Key Concept

Binary choices and the use of permutations in assignment problems

Explanation

Each problem represents a binary choice for assignment to a student, and since we exclude the case where no problems are assigned, we subtract one from the total number of binary combinations. If there are multiple students, this number is raised to the power of the number of students to find the total number of assignments.

I have to overcome 5 problems to do among 20 students, so each student may have to do more than one problem. What is the likelihood that a student will have more than one problem? You need to give the result as a rounded real number to 3 decimal places, separating the integer part from the decimal part with a dot (not a comma).

Solution by Steps

step 1

To determine the likelihood that a student will have more than one problem, we first calculate the probability that a student will have exactly one problem. Since there are 5 problems to distribute among 20 students, we can use the binomial coefficient to find the number of ways to assign one problem to one student

step 2

The number of ways to assign one problem to one student is given by the binomial coefficient

\binom{20}{1}

, which is simply 20

step 3

The total number of ways to distribute 5 problems among 20 students without any restrictions is

\binom{20}{5}

step 4

Calculate the probability

P

that a student gets exactly one problem by dividing the number of ways to give one student one problem by the total number of ways to distribute the problems:

P = \frac{\binom{20}{1}}{\binom{20}{5}}

step 5

Compute the probability using the binomial coefficients:

P = \frac{20}{\frac{20!}{5!(20-5)!}} = \frac{20}{\frac{20!}{5!15!}}

step 6

Simplify the expression:

P = \frac{20 \times 5! \times 15!}{20!} = \frac{20 \times 5 \times 4 \times 3 \times 2 \times 1}{20 \times 19 \times 18 \times 17 \times 16}

step 7

Cancel out the common factors and compute the probability:

P = \frac{5 \times 4 \times 3 \times 2 \times 1}{19 \times 18 \times 17 \times 16} = \frac{120}{146,880}

step 8

The probability that a student gets exactly one problem is:

P = \frac{120}{146,880} \approx 0.000817

step 9

To find the probability that a student gets more than one problem, we subtract the probability of getting exactly one problem from 1:

P(\text{more than one}) = 1 - P(\text{exactly one})

step 10

Calculate the final probability:

P(\text{more than one}) = 1 - 0.000817

step 11

Simplify and round to three decimal places:

P(\text{more than one}) \approx 1 - 0.000817 = 0.999183

Answer

0.999

Key Concept

Probability of a student getting more than one problem

Explanation

The probability that a student will have more than one problem is calculated by subtracting the probability of a student getting exactly one problem from 1. The result is rounded to three decimal places as requested.

Un test de detecció precoç d'una malaltia dóna positiu el 98% de les vegades en què existeix la malaltia, i un 8.5% de les vegades en què no existeix la malaltia. Suposem que un 0.8% de la població té aquesta malaltia.. Si aplicam el test a un individu i dóna positiu, quina és la probabilitat que aquest individu tengui la malaltia? Heu de donar el resultat exacte com un nombre real arrodonit a 3 xifres decimals, separant la part entera de la part decimal amb un punt (no amb una coma).

Solution by Steps

step 1

Define the probabilities given:

P(D)

is the probability of having the disease,

P(\overline{D})

is the probability of not having the disease,

P(+|D)

is the probability of a positive test given the disease, and

P(+|\overline{D})

is the probability of a positive test given no disease

step 2

Use Bayes' theorem to find the probability of having the disease given a positive test result:

P(D|+) = \frac{P(+|D) \cdot P(D)}{P(+|D) \cdot P(D) + P(+|\overline{D}) \cdot P(\overline{D})}

step 3

Substitute the given probabilities into Bayes' theorem:

P(D|+) = \frac{0.98 \cdot 0.008}{0.98 \cdot 0.008 + 0.085 \cdot (1 - 0.008)}

step 4

Calculate the probability:

P(D|+) = \frac{0.98 \cdot 0.008}{0.98 \cdot 0.008 + 0.085 \cdot 0.992}

step 5

Perform the arithmetic to find the exact probability:

P(D|+) = \frac{0.00784}{0.00784 + 0.08432}

step 6

Simplify the expression to get the final probability:

P(D|+) = \frac{0.00784}{0.09216}

step 7

Calculate the final probability and round to three decimal places:

P(D|+) \approx 0.085

Answer

0.085

Key Concept

Bayes' theorem

Explanation

Bayes' theorem is used to find the probability of an event based on prior knowledge of conditions that might be related to the event. In this case, it helps us calculate the probability that an individual has the disease given a positive test result.

Un test de detecció d'una malaltia té un coeficient de falsos positius de 0.25 i un coeficient de falsos negatius de 0.08. Si un 12% de la població té aquesta malaltia, quina és la probabilitat que el test, efectuat sobre un individu escollit a l'atzar, doni positiu? Heu de donar el resultat exacte com un nombre decimal, separant la part entera de la part decimal amb un punt (no amb una coma).

Solution by Steps

step 1

Calculate the probability of a true positive. This is the probability that a person has the disease and the test correctly identifies it as positive. Given that 12% of the population has the disease, the probability of a true positive is the same as the prevalence of the disease:

P(\text{Disease}) = 0.12

step 2

Calculate the probability of a false positive. This is the probability that a person does not have the disease but the test incorrectly identifies it as positive. The false positive rate is 0.25, and since 88% of the population does not have the disease:

P(\text{False Positive}) = 0.25 \times (1 - P(\text{Disease})) = 0.25 \times 0.88

step 3

Add the probabilities of a true positive and a false positive to find the total probability that the test will give a positive result.

P(\text{Positive}) = P(\text{Disease}) + P(\text{False Positive})

step 4

Perform the calculations to find the exact probability as a decimal.

P(\text{Positive}) = 0.12 + (0.25 \times 0.88)

P(\text{Positive}) = 0.12 + 0.22

P(\text{Positive}) = 0.34

Answer

The probability that the test, when performed on a randomly chosen individual, will give a positive result is 0.34.

Key Concept

Total probability of a positive test result

Explanation

The total probability of a positive test result is the sum of the probabilities of a true positive and a false positive.

Siguin A i B esdeveniments tals que P(A)=0.4, P(B)=0.6 i P(A\cap B)=0.2. Són A i B independents? Heu de contestar ``si'' o ``no'' (sense cometes, i sense accent).

Solution by Steps

step 1

To determine if events A and B are independent, we need to check if the product of their individual probabilities equals the probability of their intersection:

P(A) \cdot P(B) = P(A \cap B)

step 2

Calculate the product of

P(A)

and

P(B)

:

P(A) \cdot P(B) = 0.4 \cdot 0.6

step 3

Perform the multiplication:

0.4 \cdot 0.6 = 0.24

step 4

Compare the result from step 3 with

P(A \cap B)

:

0.24 \neq 0.2

Answer

no

Key Concept

Independence of events

Explanation

Events A and B are independent if and only if

P(A) \cdot P(B) = P(A \cap B)

. Since

0.24 \neq 0.2

, A and B are not independent.

D'un total de 523 rosers d'un jardí, 80 estan infectats amb fongs i 63 estan infestats de mosca blanca, i 6 tenen les dues malalties. Quants rosers d'aquest jardí no pateixen cap de les dues plagues?

Solution by Steps

step 1

To calculate the number of healthy rose bushes, we need to apply the principle of inclusion-exclusion. The formula is:

Total = A + B - (A \cap B)

where

A

is the number of rose bushes infected with fungi,

B

is the number of rose bushes infested with whiteflies, and

A \cap B

is the number of rose bushes with both diseases

step 2

Substitute the given values into the formula:

Total = 80 + 63 - 6

step 3

Perform the arithmetic operation:

Total = 80 + 63 - 6 = 137

This is the total number of rose bushes affected by at least one of the diseases

step 4

Subtract the number of affected rose bushes from the total number of rose bushes to find the number of healthy ones:

Healthy = 523 - 137

step 5

Perform the subtraction:

Healthy = 523 - 137 = 386

Answer

386

Key Concept

Inclusion-Exclusion Principle

Explanation

The inclusion-exclusion principle is used to avoid double-counting the rose bushes that have both diseases when calculating the total number of affected bushes.

Un test de detecció d'una malaltia té un coeficient de falsos positius de 0.2 i un coeficient de falsos negatius de 0.05. Si un 10% de la població té aquesta malaltia, quina és la probabilitat que el test, efectuat sobre un individu escollit a l'atzar, doni positiu? Heu de donar el resultat exacte com un nombre decimal, separant la part entera de la part decimal amb un punt (no amb una coma).

Solution by Steps

step 1

Calculate the probability of a true positive: The probability that an individual has the disease is 0.1, and the test correctly identifies the disease (true positive).

P(\text{Disease}) = 0.1

P(\text{True Positive}) = 1 - P(\text{False Negative}) = 1 - 0.05 = 0.95

P(\text{Positive} | \text{Disease}) = P(\text{Disease}) \times P(\text{True Positive}) = 0.1 \times 0.95

step 2

Calculate the probability of a false positive: The probability that an individual does not have the disease is 0.9, and the test incorrectly identifies the disease (false positive).

P(\text{No Disease}) = 1 - P(\text{Disease}) = 0.9

P(\text{False Positive}) = 0.2

P(\text{Positive} | \text{No Disease}) = P(\text{No Disease}) \times P(\text{False Positive}) = 0.9 \times 0.2

step 3

Add the probabilities of a true positive and a false positive to find the total probability of a positive test result.

P(\text{Positive}) = P(\text{Positive} | \text{Disease}) + P(\text{Positive} | \text{No Disease})

step 4

Perform the calculations to find the exact probability as a decimal.

P(\text{Positive}) = (0.1 \times 0.95) + (0.9 \times 0.2)

P(\text{Positive}) = 0.095 + 0.18

P(\text{Positive}) = 0.275

Answer

The probability that the test, conducted on a randomly chosen individual, will give a positive result is 0.275.

Key Concept

Total probability of a positive test result

Explanation

The total probability of a positive test result is the sum of the probabilities of a true positive and a false positive.

Un test de detecció precoç d'una malaltia dóna positiu el 98% de les vegades en què existeix la malaltia, i un 8% de les vegades en què no existeix la malaltia. Suposem que un 1.8% de la població té aquesta malaltia.. Si aplicam el test a un individu i dóna positiu, quina és la probabilitat que aquest individu tengui la malaltia? Heu de donar el resultat exacte com un nombre real arrodonit a 3 xifres decimals, separant la part entera de la part decimal amb un punt (no amb una coma).

Solution by Steps

step 1

Define the given probabilities: Sensitivity (True Positive rate)

P(\text{Positive}|\text{Disease}) = 0.98

, False Positive rate

P(\text{Positive}|\text{No Disease}) = 0.08

, and Prevalence

P(\text{Disease}) = 0.018

step 2

Calculate the probability of not having the disease

P(\text{No Disease}) = 1 - P(\text{Disease}) = 1 - 0.018 = 0.982

step 3

Use Bayes' Theorem to find the probability of having the disease given a positive test result

P(\text{Disease}|\text{Positive})

. Bayes' Theorem states:

P(\text{Disease}|\text{Positive}) = \frac{P(\text{Positive}|\text{Disease}) \cdot P(\text{Disease})}{P(\text{Positive}|\text{Disease}) \cdot P(\text{Disease}) + P(\text{Positive}|\text{No Disease}) \cdot P(\text{No Disease})}

step 4

Substitute the given values into Bayes' Theorem:

P(\text{Disease}|\text{Positive}) = \frac{0.98 \cdot 0.018}{0.98 \cdot 0.018 + 0.08 \cdot 0.982}

step 5

Perform the calculations:

P(\text{Disease}|\text{Positive}) = \frac{0.98 \cdot 0.018}{0.98 \cdot 0.018 + 0.08 \cdot 0.982} \approx \frac{0.01764}{0.01764 + 0.07856} \approx \frac{0.01764}{0.0962} \approx 0.1834

Answer

0.183

Key Concept

Bayes' Theorem

Explanation

Bayes' Theorem is used to find the probability of an event based on prior knowledge of conditions that might be related to the event. In this case, it helps us calculate the probability that an individual has the disease given a positive test result, taking into account the test's sensitivity, false positive rate, and the prevalence of the disease in the population.

Un test de detecció precoç d'una malaltia dóna positiu el 96% de les vegades en què existeix la malaltia, i un 12.5% de les vegades en què no existeix la malaltia. Suposem que un 2.8% de la població té aquesta malaltia. Si aplicam el test a un individu i dóna NEGATIU, quina és la probabilitat que aquest individu tengui la malaltia? Heu de donar el resultat exacte com un nombre real arrodonit a 5 xifres decimals, separant la part entera de la part decimal amb un punt (no amb una coma).

Solution by Steps

step 1

Define the given probabilities: Let

P(D)

be the probability of having the disease,

P(+|D)

be the probability of a positive test given the disease,

P(+|\neg D)

be the probability of a positive test given no disease, and

P(-|D)

be the probability of a negative test given the disease. We are given

P(D) = 0.028

,

P(+|D) = 0.96

, and

P(+|\neg D) = 0.125

. We need to find

P(-|D)

, which is

1 - P(+|D)

step 2

Calculate

P(-|D)

:

P(-|D) = 1 - P(+|D) = 1 - 0.96 = 0.04

step 3

Apply Bayes' theorem to find

P(D|-)

, the probability of having the disease given a negative test result. Bayes' theorem states:

P(D|-) = \frac{P(-|D) \cdot P(D)}{P(-)}

where

P(-)

is the total probability of a negative test result

step 4

Calculate

P(-)

, the total probability of a negative test result, which is the sum of the probability of a negative test result when the disease is present and when the disease is not present:

P(-) = P(-|D) \cdot P(D) + P(-|\neg D) \cdot P(\neg D)

where

P(-|\neg D) = 1 - P(+|\neg D)

and

P(\neg D) = 1 - P(D)

step 5

Calculate

P(-|\neg D)

and

P(\neg D)

:

P(-|\neg D) = 1 - P(+|\neg D) = 1 - 0.125 = 0.875

P(\neg D) = 1 - P(D) = 1 - 0.028 = 0.972

step 6

Substitute

P(-|D)

,

P(D)

,

P(-|\neg D)

, and

P(\neg D)

into the formula for

P(-)

:

P(-) = 0.04 \cdot 0.028 + 0.875 \cdot 0.972

step 7

Calculate

P(-)

:

P(-) = 0.04 \cdot 0.028 + 0.875 \cdot 0.972 = 0.00112 + 0.8502 = 0.85132

step 8

Substitute

P(-|D)

,

P(D)

, and

P(-)

into Bayes' theorem to find

P(D|-)

:

P(D|-) = \frac{0.04 \cdot 0.028}{0.85132}

step 9

Calculate

P(D|-)

:

P(D|-) = \frac{0.04 \cdot 0.028}{0.85132} \approx 0.00131

Answer

0.00131

Key Concept

Bayes' theorem and total probability

Explanation

Bayes' theorem is used to find the probability of having the disease given a negative test result, taking into account the overall prevalence of the disease and the accuracy of the test.

Siguin A i B esdeveniments tals que P(A)=0.4, P(B)=0.6 i P(A\cap B)=0.24. Són A i B independents? Heu de contestar ``si'' o ``no'' (sense cometes, i sense accent).

Solution by Steps

step 1

To determine if events A and B are independent, we need to check if the product of their individual probabilities equals the probability of their intersection:

P(A) \cdot P(B) = P(A \cap B)

step 2

Calculate the product of

P(A)

and

P(B)

:

P(A) \cdot P(B) = 0.4 \cdot 0.6 = 0.24

step 3

Compare the product from step 2 with

P(A \cap B)

:

P(A \cap B) = 0.24

Answer

si

Key Concept

Independence of events

Explanation

Since

P(A) \cdot P(B) = P(A \cap B)

, events A and B are independent.

D'un total de 100 gens estudiats, hem trobat homòlegs de 75 d'ells en el genoma humà i de 79 en el genoma de la mosca de la fruita, i de 60 en el genoma de tots dos organismes. De quants de gens no hem trobat homòlegs al genoma de cap dels dos organismes?

Solution by Steps

step 1

To solve this problem, we will use the principle of inclusion-exclusion for three sets. Let

A

be the set of genes with homologs in humans,

B

be the set of genes with homologs in fruit flies, and

C

be the set of genes with homologs in both

step 2

We are given that

|A| = 75

,

|B| = 79

, and

|C| = 60

. The principle of inclusion-exclusion states that the total number of elements in at least one of the sets is

|A \cup B| = |A| + |B| - |A \cap B|

step 3

Since

C

is the intersection of

A

and

B

, we have

|A \cap B| = |C| = 60

. Thus,

|A \cup B| = 75 + 79 - 60

step 4

Calculating the number of genes with homologs in at least one of the organisms:

|A \cup B| = 75 + 79 - 60 = 94

step 5

To find the number of genes without homologs in either organism, we subtract the number of genes with homologs in at least one organism from the total number of genes studied:

100 - |A \cup B|

step 6

Calculating the number of genes without homologs in either organism:

100 - 94 = 6

Answer

There are 6 genes without homologs in either the human or fruit fly genome.

Key Concept

Principle of Inclusion-Exclusion

Explanation

The principle of inclusion-exclusion is used to find the number of elements in the union of multiple sets, avoiding double-counting of elements in the intersections of the sets.

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