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4 years ago

Give an example of a compound event. :) - (20 pts)

Mathematics

1 answer:

Sphinxa [80]4 years ago

5 0

A compound event is one in which there is more than one possible outcome.

An experiment consists of rolling two fair dice.
(Look at the picture)
1. How many possible outcomes are there for this experiment?
2. Determine if each event is compound or simple.
(a) Rolling a 1 on the first die.
(b) Rolling an even number on the first die and an odd number on the second die.
3. What is the probability of rolling an even number on the first die and an odd number on the second die?
4. What is the probability of rolling one even number and one odd number, if it doesn't matter on which die each occurs?

1. 362. (a) simple, (b) compound
3. 9/36 = 1/4
4. 18/36 = 1/2
( / means it is a fraction)

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Translate the following words into an expression:

Readme [11.4K]

A because In math theres different Definitions for each type such as Sum=Addition Quotient= Division.

Therefore A would be your answer

Appreciate if I could get a brainlist :)

6 0

3 years ago

The probability that a person in the United States has type B+ blood is 10%. Four un-related people in the United States are se

marysya [2.9K]

Answer:

a. 0.0001

b. 0.6561

c. 0.3439

d. B. The event in part (a) is unusual because its probability is less than or equal to 0.05.

Step-by-step explanation:

a. # We are given that the probability that a person in the United States has Type B+ blood = 0.10. Also we are told that four unrelated people in the United States are selected at random.

#We have to find here the probability that all four have type B+ blood.

Since the events are independent, we have :

Probability that all four have B+ blood = 0.10 x 0.10x 0.10x0.10

= 0.0001

Therefore, the probability that all four have type B+ blood is 0.0001

b. We have to find the probability that none have B+ blood. Using the complementary law of probability we have:

Probability that blood type is not B+ = 1 - 0.10= 0.90

Therefore, the probability that none have B+ blood

= 0.90 x 0.90 x 0.90x0.90=0.6561

Therefore, the probability that none have B+ blood is 0.6561

c. We have to find the probability that at least one of the four have B+ blood.

#The probability that at least one of the four have B+ blood = 1 - Probability that none have B+ blood type

=1-0.6561=0.3439

Therefore,the probability that at least one of the four has type B+ blood is 0.3439

d. An event is considered unusual if the probability of the event is small or less than 0.05 . We note that event a is the only small probabilty and is less than 0.05.

-a is thus considered unusual(the rest are all usual events)

6 0

3 years ago

The mean of a population is 74 and the standard deviation is 16. The shape of the population is unknown. Determine the probabili

bulgar [2K]

Answer:

$P(X > 75)= 0.35402$

$P(72 < X < 75 ) = 0.2529$

$P( X < 74.7) = 0.74041$

Step-by-step explanation:

From the question we are told that

The population mean is $\mu = 74$

The population standard deviation is $\sigma = 16$

Considering question a

The sample size is n = 36

Generally the standard error of mean is mathematically represented as

$\sigma_{x} = \frac{\sigma }{\sqrt{n} }$

=> $\sigma_{x} = \frac{16}{\sqrt{36} }$

=> $\sigma_{x} = 2.67$

Generally the probability that a random sample of size 36 yielding a sample mean of 75 or more is mathematically represented as

$P(X > 75) = P( \frac{X - \mu }{ \sigma_{x}} > \frac{75 - 74}{ 2.67 } )$

$\frac{X -\mu}{\sigma } = Z (The \ standardized \ value\ of \ X )$

$P(X > 75) = P(Z > 0.3745 )$

From the z table the area under the normal curve representing 0.3745 to the right is

$P(Z > 0.3745 ) = 0.35402$

=> $P(X > 75)= 0.35402$

Considering question b

The sample size is n = 104

Generally the standard error of mean is mathematically represented as

$\sigma_{x} = \frac{\sigma }{\sqrt{n} }$

=> $\sigma_{x} = \frac{16}{\sqrt{104} }$

=> $\sigma_{x} = 1.5689$

Generally the probability that a random sample of size 104 yielding a sample mean between 72 and 75 is mathematically represented as

$P(72 < X < 75 ) = P(\frac{72 - 74 }{1.5689} < \frac{X - \mu }{\sigma_{x}} < \frac{75 - 74 }{1.5689} )$

=> $P(72 < X < 75 ) = P(-1.275 < Z < 0.375 )$

=> $P(72 < X < 75 ) = P(Z < 0.375 ) - P(Z < -1.275)$

From the z table the area under the normal curve representing -1.275 to to the left is

$P(Z < -1.275) =0.10115$

=> $P(72 < X < 75 ) = 0.35402 - 0.10115$

=> $P(72 < X < 75 ) = 0.2529$

Considering question c

The sample size is n = 217

Generally the standard error of mean is mathematically represented as

$\sigma_{x} = \frac{\sigma }{\sqrt{n} }$

=> $\sigma_{x} = \frac{16}{\sqrt{217} }$

=> $\sigma_{x} = 1.086$

Generally the probability that a random sample of size 217 yielding a sample mean of less than 74.7 is mathematically represented as

$P( X < 74.7) = P(\frac{X - \mu }{\sigma_x} < \frac{ 74.7 - 74 }{ 1.086 })$

=> $P( X < 74.7) = P(Z < 0.6446 )$

From the z table the area under the normal curve representing 0.6446 to to the left is

$P(Z < 0.6446 ) = 0.74041$

=> $P( X < 74.7) = 0.74041$

8 0

3 years ago

A bag contains 120 marbles Some are red and the rest are black. There are 19 red marbles for every black marble. How many black

kobusy [5.1K]

Answer:

6 black

114 red

Step-by-step explanation:

19x6=114

6×1=6

114+6=120

7 0

4 years ago

Carissa earns $450 per week for a 40 hour work week plus $16.83 per hour for each hour of overtime after 40 hours. Write an equa

Aleksandr-060686 [28]

Answer:

Step-by-step explanation:

Easy,

$%*%%E#$(

5 0

3 years ago