Archive for the ‘probability’ Category

Problem 2005

This exercise uses the population growth model.

A grey squirrel population was introduced in  a certain county of  Great Britain 35 years ago. Biologists observe that the population doubles every 7 years, and now the population is 100,000.

What was the initial size of the squirrel population?
Estimate the squirrel population 10 years from now.

 

Solution:-

 

The initial size of the squirrel population is 3125 .

Estimate the squirrel population 10 years from  is 269,180.

 

Problem 2003

Graph the function, not by plotting points, but by starting from the graphs below.

f(x) = 3-x

g1

State the domain and range.

State the asymptote.

 

Solution:-

 

State the domain and range.

Domain (-∞,∞)

Range (0,∞)

State the asymptote.

y = 0

 

Problem 2002

Assume that SAT scores are normally distributed with mean 1518 and standard deviation 325. If one SAT score is randomly selected, find the probability that it is greater than 1600.

 

Solution:-

 

z = \frac{x-\mu }{\sigma }

=\frac{1600 - 1518}{325}

=0.25

P(z > 0.25) = 1 – 0.5987

=0.4013

 

Problem 2000

A bag has 4 green marbles, 3 red marbles, and 3 yellow marbles. What is the probability that you pick a green marble, do not replace it, and pick a red marble?

 

Solution:-

 

P(A and B) = P(A) • P(B following A)

\frac{2}{15}

 

Problem 1199

Use this table to answer the question that follows.

 

Students who Studied
Yes No Total
Students who Earned an A on the Test Yes 6 1 7
No 24 3 27
Total 30 4 34

 

 

Are the events, earn an A and study, independent or dependent?  What is the probability of randomly selecting a student who did not earn an A given that the student studied?  Express the answer as a fraction in simplest form.

 

Solution:-

 

Because the probability of selecting a student who earned an A is not the same as the probability of selecting a student who earned an A given that they studied, the events are dependent.  The probability of randomly selecting a student who did not earn an A given that the student studied is

\frac{24}{30} = \frac{4}{5}

 

 

Problem 1198

Use the information in this table to find the indicated probability.

  Blood Type
A B AB O Total
Rh Factor Positive 264 63 25 277 629
Negative 47 11 4 49 111
Total 311 74 29 326 740

 


What is the probability of selecting a donor that is blood type A or is Rh-positive?  Express the answer as a decimal rounded to the nearest thousandth.

 

Solution:-

 

The events are NOT mutually exclusive, so the probability is found by adding the individual probabilities and subtracting the common probability.

 

\frac{311}{740}+\frac{629}{740}-\frac{264}{740} = \frac{676}{740}\approx 0.914

 

Problem 1197

Think about a standard deck of 52 playing cards. Which of the following events is mutually exclusive?

 

Solution:-

 

Drawing a king or a queen is mutually exclusive because the events cannot happen at the same time.

 

Problem 1196

Think about a six-sided die with each number, 1 through 6, appearing once on each die. Which of the following events is mutually inclusive?

 

Solution:-

 

Rolling a 3 or a factor of  6.

 

Problem 1168

Find P(B∩E) directly from the table.

  A B C Total
D 0.13 0.02 0.05 0.20
E 0.37 0.28 0.15 0.80
Total 0.50 0.30 0.20 1.00

 

Solution:-

 

If F and G are two events in sample space S, than the union of F and G, denoted by FUG, is define to be FUG = {e ϵS | e ϵ F or e ϵ G}, and the intersection of F and G, denoted by F∩G, is defined to be F∩G = {e ϵS | e ϵ F or e ϵ G}. Furthermore, FUG is also define to be the event F or G, while F∩G is also define to be the event F and G.

The event B∩E is define to be the event B and E.

The value of P(B∩E) can be found where the column containing B and the row containing E intersect.

Find P(B∩E) directly from the table.

P(B∩E) = 0.28

 

 

Problem 1167

A pair dice is rolled. What is the probability of getting a sum of 8?

 

Solution:-

 

To find the probability of an event, denoted P(E), you need to find the number of possible outcomes of the experiment, n(s) . This is the same as the number if elements in the sample space. You also need to find the number of ways the event can occur, denoted n(E).

Then use the formula p(E) = \frac{n(E)}{n(S)}.

The experiment in this case is the rolling of two dice.

Sum= 8({2,6},{3,5},{4,4},{5,3},{6,2})

By counting all possible outcomes it is clear that n(s) = 36.

For the event of getting a sum of 8, the number of possible outcomes is 5.

Thus n(E) = 5.

Therefore, P(E) = \frac{n(E)}{(S)} = \frac{5}{36}.