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6.4 Independent and Dependent events
If A and B are independent events, then probability of both occurring is:
P(A) x (B)
P(A) x (B)
 | Example: P(A n B) = 1/6 x 4/52 = 1/78What is the probability of rolling a 3 with a dice, and drawing a 3 from a deck of cards? Solution: |
| If event B is dependent upon even A then: P(A and B) = P(A) x P(B | A) Conditional Probability: P(A | B) = P(A n B) ÷ P(B) = |  |
The second event is conditional upon the first event's result:
(B given A has already occurred)
(B given A has already occurred)
 | Example: A family has three children. What is the probability that the family has 2 boys and a girl given the middle child is a boy? Solution: A. 2 boys and a girl B. Middle is a boy Given the formula: P(A | B) = P(A n B) ÷ P(B) = 2/8 ÷ 4/8 = 1/2 = 2/8 ÷ 4/8 = 1/2 |
Still Having trouble finding the probability for Independent Events?
Use the Product Rule.
A and B are independent of each other, which means that they do not affect each other in any way.
Independent Events
Example 1: Find the probability of rolling a six on two different dice.
1. First ask yourself, " Are these events Independent or Dependent? "
Since the outcome of one die does not affect the outcome of the other, then we
establish that these are independent events.
2. Now find the probability for each die using P(A)= n(A) ÷ n(S). You will have to
refer back to 6.1 Basic Probability Concepts.
Once the probabilities of each event is found, multiply them together to achieve the probability of A and B (rolling a 6 on two different dice).
Since, P(A n B) = P(A) x P(B) then
Probability of out take hanging and then operating system crashing afterwards: 0.01
If out take hanging, what is the probability that operating system will soon crash?
1. First ask yourself, " Are these events Independent or Dependent? "
Since the outcome of one die does not affect the outcome of the other, then we
establish that these are independent events.
2. Now find the probability for each die using P(A)= n(A) ÷ n(S). You will have to
refer back to 6.1 Basic Probability Concepts.
A(1st die) = getting a 6
B(2nd die) = getting a 6
S(1st die) = rolling anything
S(2nd die) = rolling anything
Therefore,
n(A) = 1 n(B) = 1
n(S) = 6 n(S) = 6
So,
P(A) = 1/6 P(B) = 1/6
Once the probabilities of each event is found, multiply them together to achieve the probability of A and B (rolling a 6 on two different dice).
Since, P(A n B) = P(A) x P(B) then
Conditional Probability
P(A n B)
= 1/6 x 1/6
= 1/36
Therefore the probability of rolling two dice and getting a 6 on each die is 1/36.
The conditional probability of B, P(B/A), is the probability that B occurs, given that A has already occurred.
Example 2: Out Take is an email program.
Probability of out take hanging and then operating system crashing afterwards: 0.01
If out take hanging, what is the probability that operating system will soon crash?
A = the operating system crashing
B= Out Take hanging
P(A n B) / P(B)
= P(A) / P(B)
= 0.01/0.025
= 2/5
Laurie, an avid golfer, gives herself a 70% chance of breaking par (scoring less than 72 on a round of 18 holes) if the weather is calm, but only a 15% chance of breaking par on windy days. The weather forecast gives a 40% probability of high winds tomorrow. What is the likelihood that Laurie will break par tomorrow, assuming that she plays one round of golf?
Solution:
P (par in wind or no wind) = P (par in wind) + P (par in no wind)
= (0.15)(0.4) + (0.7)(0.6)
= 0.06 + 0.42
= 0.48
Therefore, the likelihood that Laurie will break par tomorrow, assuming that she plays one round of golf is 0.48 or 48%.
Canton, Erdman, et al. Mathematics of Data Management. Toronto: McGraw-Hill Ryerson, 2002.
Page 335, Question 17.
Mrs. Richardson's notes: 6.4 Independent and Dependent Events.
antzie |
Latest page update: made by antzie
, Jun 8 2007, 2:07 AM EDT
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| Started By | Thread Subject | Replies | Last Post | ||
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| Anonymous | can you solve this question | 0 | Feb 15 2010, 4:25 PM EST by Anonymous | ||
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Thread started: Feb 15 2010, 4:25 PM EST
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What is the probability of being dealt an ace and a face card in the game “21”?
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| Anonymous | Thanks | 0 | Sep 29 2009, 7:51 PM EDT by Anonymous | ||
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Thread started: Sep 29 2009, 7:51 PM EDT
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This really helps, and the spelling really dosen't effect the understanding,
All in All, good thread. {A}{R}{K} |
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| __beckyy | ... | 2 | May 14 2008, 4:21 PM EDT by Anonymous | ||
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Thread started: Jan 10 2007, 7:31 AM EST
Watch
You might want to watch the spelling. For the explanation for the dependent events, "occuring" is spelt with two r's. And for the example of the dependent events, you guys spelt 'knowing' with two n's. For the examples and stuff for the independent/dependent events, it's all bolded and underlined---it seems clustered and is hard to read. The rest of the page looks really good and is well organized.
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