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Dmitriy789 [7]
3 years ago
5

Consider the following scenarios, where we apply probability to a game of flipping coins. In the game, we flip one coin each rou

nd. The game will not stop until two consecutive heads appear. (a) What is the probability that the game ends by flipping exactly five coins? (b) Given that the game ends after flipping the fifth coin, what is the probability that three heads appear in the sequence?
Mathematics
1 answer:
ExtremeBDS [4]3 years ago
8 0

Answer:

3.125%

0%

Step-by-step explanation:

A regular coin has two sides, meaning that each side has the same probability of landing which is 50%. Therefore, for the game to end exactly on the fifth flip it would need to land 3 times in a row on tails and the last 2 times on heads. Therefore, to get this probability we need to multiply 0.50 (50%) 5 times

0.5 * 0.5 * 0.5 * 0.5 * 0.5 = 0.03125 or 3.125%

Now the probability of three heads appearing in a sequence is 0% because after the second head in a row the game will automatically end.

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Find the area of an equilateral triangle with a side of 6 inches. Answers are:
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Answer:

x^3+3x^2+2x

Step-by-step explanation:

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Consecutive Integers: Integers that follow each other continuously in the order from smallest to largest.

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Let X denote the distance (m) that an animal moves from its birth site to the first territorial vacancy it encounters. Suppose t
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Answer:

a) P(X \leq 100) = 1- e^{-0.01342*100} =0.7387

P(X \leq 200) = 1- e^{-0.01342*200} =0.9317

P(100\leq X \leq 200) = [1- e^{-0.01342*200}]-[1- e^{-0.01342*100}] =0.1930

b) P(X>223.547) = 1-P(X\leq 223.547) = 1-[1- e^{-0.01342*223.547}]=0.0498

c) m = \frac{ln(0.5)}{-0.01342}=51.65

d) a = \frac{ln(0.05)}{-0.01342}=223.23

Step-by-step explanation:

Previous  concepts

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:

P(X=x)=\lambda e^{-\lambda x}

Solution to the problem

For this case we have that X is represented by the following distribution:

X\sim Exp (\lambda=0.01342)

Is important to remember that th cumulative distribution for X is given by:

F(X) =P(X \leq x) = 1-e^{-\lambda x}

Part a

For this case we want this probability:

P(X \leq 100)

And using the cumulative distribution function we have this:

P(X \leq 100) = 1- e^{-0.01342*100} =0.7387

P(X \leq 200) = 1- e^{-0.01342*200} =0.9317

P(100\leq X \leq 200) = [1- e^{-0.01342*200}]-[1- e^{-0.01342*100}] =0.1930

Part b

Since we want the probability that the man exceeds the mean by more than 2 deviations

For this case the mean is given by:

\mu = \frac{1}{\lambda}=\frac{1}{0.01342}= 74.516

And by properties the deviation is the same value \sigma = 74.516

So then 2 deviations correspond to 2*74.516=149.03

And the want this probability:

P(X > 74.516+149.03) = P(X>223.547)

And we can find this probability using the complement rule:

P(X>223.547) = 1-P(X\leq 223.547) = 1-[1- e^{-0.01342*223.547}]=0.0498

Part c

For the median we need to find a value of m such that:

P(X \leq m) = 0.5

If we use the cumulative distribution function we got:

1-e^{-0.01342 m} =0.5

And if we solve for m we got this:

0.5 = e^{-0.01342 m}

If we apply natural log on both sides we got:

ln(0.5) = -0.01342 m

m = \frac{ln(0.5)}{-0.01342}=51.65

Part d

For this case we have this equation:

P(X\leq a) = 0.95

If we apply the cumulative distribution function we got:

1-e^{-0.01342*a} =0.95

If w solve for a we can do this:

0.05= e^{-0.01342 a}

Using natural log on btoh sides we got:

ln(0.05) = -0.01342 a

a = \frac{ln(0.05)}{-0.01342}=223.23

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