Katie C.

# If the animal is initially in the woods, what is the probability that it is in the meadow on the next three observations?

If the animal is in the woods on one observation, then it is three times as likely to be in the woods as the meadows on the next observation. If the animal is in the meadows on one observation, then it is twice as likely to be in the meadows as the woods on the next observation.

Assume that state 1 is being in the meadows and that state 2 is being in the woods.

1.If the animal is initially in the woods, what is the probability that it is in the woods on the next three observations?

2.If the animal is initially in the woods, what is the probability that it is in the meadow on the next three observations?

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Katie C.

Thank you!! could you help me with these problems as well?

1.Assume that producing 1 unit of coal requires 0.4 units of coal, 0.1 units of lumber, and 0.2 units of steel; that producing 1 unit of lumber requires 0 units of coal, 0.7 units of lumber, and 0.2 units of steel; and that producing 1 unit of steel requires 0.8 units of coal, 0.2 units of lumber, and 0.5 units of steel.

Find the production schedule that satisfies an external demand for 43 units of coal, 26 units of lumber, and 17 units of steel.

2.Find the equations of the lines through the following pairs of points, in slope intercept form.
(1) Find the equation of the line through (5,3),(−4,4).
y=

(2) Find the equation of the line through (2,−2),(5,3).
y=

3. If she made the last free throw, then her probability of making the next one is 0.9. On the other hand, If she missed the last free throw, then her probability of making the next one is 0.5.

Assume that state 1 is Makes the Free Throw and that state 2 is Misses the Free Throw.

-If she makes her first free throw, what is the probability that she makes the third one?

-If she misses her first free throw, what is the probability that she makes the third one?

Report

12/15/15

Caius L.

1. I don't know what a production schedule is referring to.
2. To find the slope of a line using two points using the equation y2-y/ (x2 - x1). In words, this is simply change in y divided by change in x.

So slope m = (4-3)/(-4-5) = -1/9   (notice how when you graph the two points, you can clearly see how there is a movement of 1 down in the vertical direction for a movement of 9 over in the right direction. This clearly corresponds to -1  vertical PER every 9 horizontal. A good thing to gain an intuition for...lines come up a lot in math...

To find the equation of the line use the equation y - y= m(x - x1). In this equation, x and y are simply the variables that will stay as x and y all the way through the final simplification of the equation and x1 and y1 are where we can substitute the x and y coordinates of either of the points, either one of them will work since we already know the slope that connects them. m is traditionally used as the letter denoting the slope. Say we use the point (5,3) since it has positive values and is easier to work with:

y - 3 = -1/9(x - 5)

This is already the equation of the line, but slope intercept form means we must isolate the y on one side.
y = -1/9x + 5/9 + 3
y = -1/9x + 32/9

When in this form the equation clearly shows us how the slope of the line is -1/9 and the y-intercept of the line is 32/9.

The other line problem is the exact same idea.

3a) This is very similar to the animal problem but we need to consider both possibilities for what happens for the second free throw. For example, for the first scenario there are two ways that she makes the third free throw: either she makes the second and then she makes the third OR she misses the second and makes the third. We need to figure out the possibilities for these two scenarios and add them together.

Scenario 1, she makes the second: since the makes the first, there is a 9/10 probability of making the 2nd. Then there is a 9/10 probability of making the 3rd. Multiplying we get 81/100 for making the third one this way.

Scenario 2, she misses the second: since she made first, there is a 1/10 probability of missing the 2nd. Then there is a 5/10 probability of making the 3rd. Multiplying you get 1/10 * 5/10 = 5/100 for making the third one this way.

Adding both of these together since they are separate paths to making the 3rd freethrow you get: 81/100 + 5/100 = 86/100 = 86%

3b)-If she misses her first free throw, what is the probability that she makes the third one?
Again, there are two "paths" to making the 3rd freethrow: either making the 2nd and then using the appropriate probability or missing the 2nd and using the appropriate probability for that scenario.
making the 2nd is a probability of 5/10 and then making the 3rd 9/10 so altogether 5/10*9/10 = 45/100
missing the second has a probability of 5/10 and then making the 3rd 5/10 so altogether 25/100.
Adding these up: 45/100+25/100 = 70/100 = 70%
Report

12/15/15

Katie C.

Thank you again I just have 1 more question!!

1. volume of securities sold by a brokerage firm, using the following data:

(i) If volume is high this week, then next week it will be high with a probability of 0.9 and low with a probability of 0.1.
(ii) If volume is low this week then it will be high next week with a probability of 0.3.

The manager estimates that the volume is five times as likely to be high as to be low this week.

Assume that state 1 is high volume and that state 2 is low volume.

(1) Find the transition matrix P for this Markov chain:

(2) Find the state vector that represents the manager's estimate

(3) Using this estimate as the initial state vector, find the state vector for two weeks from now:

What is the probability that two weeks from now the volume will be high?

(4) Again, using the manager's estimate as the initial state vector, find the state vector for three weeks from now:

What is the probability that three weeks from now the volume will be high?

(5) Suppose, contrary to the manager's estimate, that this week the volume is low. How many weeks must pass before a week comes along in which the probability of high volume is at least 0.6?

Report

12/16/15

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