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2 years ago

Give an example of an event that is certain and an event that is impossible. What are the probabilities for each event?

Mathematics

2 answers:

bixtya [17]2 years ago

8 0

Answer:

impossible event

people cant live without water

getting both a head and a tail when flipping a coin

obtaining seven on a throw of a die

certain event

it is 2pm now and will be 4pm after two hours

if you are sixteen,the likelihood of your turning seventeen on you next birthday

if it is thursday the probability that tommorow is friday An event that is certain is something that will definitely happen, such as the sun will rise tomorrow, or when rolling a six sided die, it will land on 1 - 6.

Something impossible, includes events that will never happen, such as pig will fly, or when rolling a six sided die, it will land on 7.

2. The experimental probability may be different to the theoretical probability as theoretical probability is only theory, and when trying something out in real life, there is a high chance that it will be different.

Step-by-step explanation:

There is question in my book: Probability of impossible event is?

After reading the question my instant answer was 0 and that was the answer given.

But then i thought other way, question is probability of impossible event, so there are two outcomes possible or impossible (event can be certain or impossible).

Therefore probability of impossible outcome is 12.

Can that also be answer?

If this doesn't help Get Tutor Immediately!

OleMash [197]2 years ago

3 0

Answer:

Thank you for posting your question here at brainly. Feel free to ask more questions.

The best and most correct answer among the choices provided by the question is 0 (zero).

Hope my answer would be a great help for you. 3 to all sorry i can't help you all i try my best tho my goal is to get to 4 brainlyest may 8 if i'm lucky god bless you all stay safe during these tuff times

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What is the solution to the equation below? Round your answer to two

noname [10]

Answer:

D. x=5

The answer is x=4.50 then it is rounded off to 5. So, the answer is 5.

Step-by-step explanation:

Find the domain of the inequality: $2x-1\geq 0$

Rearrange terms to the left side of the equation: $2x\geq 1$

Divide both sides of the inequality by the coefficient of the variable:

=> $x\geq \frac{1}{2}$

The domain of the inequality is: $x\geq \frac{1}{2}$

Convert logarithm to exponential form: $2^{3}=2x-1$

Calculate the power: $8=2x-1$

Rearrange terms to the left side of the equation: $-2x= -8-1$

Calculate: $-2x=-9$

Divide both sides of the equation by the coefficient of the variable: $x=\frac{-9}{-2}$

=> $x=\frac{9}{2}$

=> $x=4.5\\$

Round the number: $x=4.50$

Answer: $x=4.50$

=> $x=5$

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2 years ago

What trig function would I use to solve for x?

IRINA_888 [86]

Answer:

Cosine

Step-by-step explanation:

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2 years ago

Write an addition sentence that has a sun of 20

Vlada [557]

10+10=20, 15+5=20 Those are both addition sentences that have a sum of 20.

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4 years ago

Read 2 more answers

Twenty percent of drivers driving between 10 pm and 3 am are drunken drivers. In a random sample of 12 drivers driving between 1

Lesechka [4]

Answer:

(a) 0.28347

(b) 0.36909

(d) 0.9806

Step-by-step explanation:

Given information:

n=12

p = 20% = 0.2

q = 1-p = 1-0.2 = 0.8

Binomial formula:

$P(x=r)=^nC_rp^rq^{n-r}$

(a) Exactly two will be drunken drivers.

$P(x=2)=^{12}C_{2}(0.2)^{2}(0.8)^{12-2}$

$P(x=2)=66(0.2)^{2}(0.8)^{10}$

$P(x=2)=\approx 0.28347$

Therefore, the probability that exactly two will be drunken drivers is 0.28347.

(b)Three or four will be drunken drivers.

$P(x=3\text{ or }x=4)=P(x=3)\cup P(x=4)$

$P(x=3\text{ or }x=4)=P(x=3)+P(x=4)$

Using binomial we get

$P(x=3\text{ or }x=4)=^{12}C_{3}(0.2)^{3}(0.8)^{12-3}+^{12}C_{4}(0.2)^{4}(0.8)^{12-4}$

$P(x=3\text{ or }x=4)=0.236223+0.132876$

$P(x=3\text{ or }x=4)\approx 0.369099$

Therefore, the probability that three or four will be drunken drivers is 0.3691.

(c)

At least 7 will be drunken drivers.

$P(x\geq 7)=1-P(x$

$P(x\leq 7)=1-[P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4)+P(x=5)+P(x=6)]$

$P(x\leq 7)=1-[0.06872+0.20616+0.28347+0.23622+0.13288+0.05315+0.0155]$

$P(x\leq 7)=1-[0.9961]$

$P(x\leq 7)=0.0039$

Therefore, the probability of at least 7 will be drunken drivers is 0.0039.

(d) At most 5 will be drunken drivers.

$P(x\leq 5)=P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4)+P(x=5)$

$P(x\leq 5)=0.06872+0.20616+0.28347+0.23622+0.13288+0.05315$

$P(x\leq 5)=0.9806$

Therefore, the probability of at most 5 will be drunken drivers is 0.9806.

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3 years ago

Roberto wants to estimate the answer to this problem to see if his answer is reasonable.

marusya05 [52]

5849 rounded to the nearest hundred
- The number '8' is located in the hundreds place
- Since there is a '4' in the tens place, it's telling number '8' to stay the same
5849 ⇒ 5800

2621 rounded to the nearest hundred
- The number '6' is located in the hundreds place
- Since there is a number '2' in the tens place, it's telling number '6' to stay the same
2621 ⇒ 2600

Answer: 5800 - 2600

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3 years ago