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3 years ago

Four coins are tossed. What is the probability of at least one tail?

Mathematics

2 answers:

xxTIMURxx [149]3 years ago

5 0

Step-by-step explanation:

I can't understand it so plz don't mind or else I will mind

sladkih [1.3K]3 years ago

4 0

Answer:

93.75%

This type of problem you have to REVERSE the question...

"at least one tail" is all possibilities EXCEPT "all heads"...

so your answer is 100% - P(4 heads)

P(4 heads) = (1/2)^4 = .0625

1 - .0625 = .9375

Step-by-step explanation:

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Question 1 of 29

Zina [86]

The characteristic of the repeating shapes in a tessellation is that shapes cannot have spaces between them.

<h3>What is tessellation ?</h3>

When we talk about tessellation, our minds go to a situation in which many shapes especially polygons are packed together without spaces between them.

Thus, the characteristic of the repeating shapes in a tessellation is that shapes cannot have spaces between them.

Learn more about tessellation: brainly.com/question/3294818

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8 0

2 years ago

The number of defective circuit boards coming off a soldering machine follows a Poisson distribution. During a specific ten-hour

Alexus [3.1K]

Answer:

a) the probability that the defective board was produced during the first hour of operation is $\frac{1}{10}$ or 0.1000

b) the probability that the defective board was produced during the last hour of operation is $\frac{1}{10}$ or 0.1000

c) the required probability is 0.2000

Step-by-step explanation:

Given the data in the question;

During a specific ten-hour period, one defective circuit board was found.

Lets X represent the number of defective circuit boards coming out of the machine , following Poisson distribution on a particular 10-hours workday which one defective board was found.

Also let Y represent the event of producing one defective circuit board, Y is uniformly distributed over ( 0, 10 ) intervals.

f(y) = $\left \{ {{\frac{1}{b-a} }\\\ }} \right _0$ ; ( a ≤ y ≤ b ) $_{elsewhere$

= $\left \{ {{\frac{1}{10-0} }\\\ }} \right _0$ ; ( 0 ≤ y ≤ 10 ) $_{elsewhere$

f(y) = $\left \{ {{\frac{1}{10} }\\\ }} \right _0$ ; ( 0 ≤ y ≤ 10 ) $_{elsewhere$

Now,

a) the probability that it was produced during the first hour of operation during that period;

P( Y < 1 ) = $\int\limits^1_0 {f(y)} \, dy$

we substitute

= $\int\limits^1_0 {\frac{1}{10} } \, dy$

= $\frac{1}{10} [y]^1_0$

= $\frac{1}{10} [ 1 - 0 ]$

= $\frac{1}{10}$ or 0.1000

Therefore, the probability that the defective board was produced during the first hour of operation is $\frac{1}{10}$ or 0.1000

b) The probability that it was produced during the last hour of operation during that period.

P( Y > 9 ) = $\int\limits^{10}_9 {f(y)} \, dy$

we substitute

= $\int\limits^{10}_9 {\frac{1}{10} } \, dy$

= $\frac{1}{10} [y]^{10}_9$

= $\frac{1}{10} [ 10 - 9 ]$

= $\frac{1}{10}$ or 0.1000

Therefore, the probability that the defective board was produced during the last hour of operation is $\frac{1}{10}$ or 0.1000

no defective circuit boards were produced during the first five hours of operation.

probability that the defective board was manufactured during the sixth hour will be;

P( 5 < Y < 6 | Y > 5 ) = P[ ( 5 < Y < 6 ) ∩ ( Y > 5 ) ] / P( Y > 5 )

= P( 5 < Y < 6 ) / P( Y > 5 )

we substitute

$= (\int\limits^{6}_5 {\frac{1}{10} } \, dy) / (\int\limits^{10}_5 {\frac{1}{10} } \, dy)$

$= (\frac{1}{10} [y]^{6}_5) / (\frac{1}{10} [y]^{10}_5)$

= ( 6-5 ) / ( 10 - 5 )

= 0.2000

Therefore, the required probability is 0.2000

4 0

3 years ago

Even if a treatment has an effect, it is still possible to obtain a sample mean after the treatment that is very similar to the

vlada-n [284]

Answer:

Step-by-step explanation:

Hello!

Suppose that the objective of the experiment is to test if a certain treatment modifies the mean of the population of interest.

If for example, the treatment is "new fertilizer" and the population of interest is "yield of wheat crops"

Then you'd expect that using the new fertilizer will at least modify the average yield of the wheat crops.

The hypotheses will be then

H₀: μ = μ₀

H₁: μ ≠ μ₀

Where μ₀ represents the known average yield of wheat crops. (is a value, for this exercise purpose there is no need to know it)

We know that the treatment modifies the population mean, i.e. the null hypothesis is false.

The sample we took to test whether or nor the new fertilizer works conducts us to believe, it does not affect, in other words, we fail to reject the null hypothesis.

Then we are in a situation where we failed to reject a false null hypothesis, this situation is known as <em><u>Type II error</u></em>.

I hope this helps!

4 0

4 years ago

Find the measure of an exterior angle of a regular polygon with 9 sides

Contact [7]

Answer: 40°

Step-by-step explanation: The formula for calculating the size of an exterior angle is: exterior angle of a polygon = 360 ÷ number of sides. So it's 360°÷9=40°

3 0

3 years ago

Plzzzz help me with this 20 points

nydimaria [60]

the answer is there is only 1 solution

6 0

3 years ago

Read 2 more answers

Four coins are tossed. What is the probability of at least one tail?​

Four coins are tossed. What is the probability of at least one tail?