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solve a word problem

by Susan
(LA,CA)











































The probability that San Francisco
plays in the next Super Bowl is nine times the probability
that they do not play in the next Super Bowl. The probability
that San Francisco plays in the next Super Bowl plus the
probability that they do not play is 1. What is the probability
that San Francisco plays in the next Super Bowl

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Jan 04, 2011
Probability Word Problem
by: Staff

The question:

by Susan
(LA, CA)




The probability that San Francisco
plays in the next Super Bowl is nine times the probability
that they do not play in the next Super Bowl. The probability
that San Francisco plays in the next Super Bowl plus the
probability that they do not play is 1. What is the probability
that San Francisco plays in the next Super Bowl?

The answer:

This is a classic problem involving two equations and two unknowns.


For this problem, I am going to define the two unknowns as follows:

P(Play) = the probability that San Francisco will play in the next Super Bowl

P(Not Play) = the probability that San Francisco will not play in the next Super Bowl


The two equations can be written as follows:

P(Play) + P(Not Play) = 1

P(Play) = 9 * P(Not Play)

I am going to use substitution to solve for the two unknowns. However, you can use the addition method just as well. In either case the answer will be the same.

Since P(Play) = 9 * P(Not Play), we can substitute the 9 * P(Not Play) for P(Play) . . . or vice versa.

I will now replace P(Play) in the first equation with 9 * P(Not Play).

P(Play) + P(Not Play) = 1

9 * P(Not Play) + P(Not Play) = 1


Solving for P(Not Play)


Combine Like Terms

9 * P(Not Play) + P(Not Play) = 1

10 * P(Not Play) = 1

Divide each side of the equation by 10 to eliminate the 10 on the left side of the equation

10 * P(Not Play) = 1

[10 * P(Not Play)]/10 = 1/10

(10/10) * P(Not Play) = 1/10

(1) * P(Not Play) = 1/10

P(Not Play) = 1/10, or 0.1

Solve for P(Play)

P(Play) + P(Not Play) = 1

Since P(Not Play) = .1, the value of .1 can be substituted for P(Not Play)

P(Play) + P(Not Play) = 1

P(Play) + .1 = 1


Subtract the .1 from each side of the equation

P(Play) + .1 = 1

P(Play) + .1 - .1 = 1 - .1

P(Play) + 0 = .9

P(Play) = .9

To convert the .9 to a percent, multiply by 100

P(Play) = 90%



The final answer is: P(Play) = 90%. The probability the San Francisco will play in the Super Bowl is 90%.





Thanks for writing.


Staff
www.solving-math-problems.com


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