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− | [[Category:ECE302Spring2013Boutin]] [[Category:ECE]] [[Category:ECE302]] [[Category:probability]] [[Category | + | [[Category:ECE302Spring2013Boutin]] [[Category:ECE]] [[Category:ECE302]] [[Category:probability]] [[Category:problem solving]] |
[[Category:conditional probability]] | [[Category:conditional probability]] | ||
− | + | ==Problem Involving Conditional Probability== | |
+ | Question: | ||
+ | |||
If there are 2 white balls and 4 red balls in box 1, and 5 white balls and 3red balls in box 2. Now randomly take one ball from box 1 and put into box 2. Then take one ball from box 2. | If there are 2 white balls and 4 red balls in box 1, and 5 white balls and 3red balls in box 2. Now randomly take one ball from box 1 and put into box 2. Then take one ball from box 2. | ||
− | + | # If under the condition that the ball being taken from box1 is red, what is the possibility that the ball that's taken from box 2 is also red? | |
− | + | # What's the possibility that the ball that's taken from box 2 is red? | |
+ | ---- | ||
+ | Answer: | ||
− | |||
(1)Assume event A is the ball that's taken from box 2 is red; and event B is the ball that's taken from box 1 is red. | (1)Assume event A is the ball that's taken from box 2 is red; and event B is the ball that's taken from box 1 is red. | ||
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=(4/9)*(2/3) + (1/3)*(1/3) | =(4/9)*(2/3) + (1/3)*(1/3) | ||
=11/27 | =11/27 | ||
+ | ---- | ||
+ | ==Comments/discussion== | ||
+ | *Write comment here | ||
+ | **answer here | ||
[[Bonus_point_1_ECE302_Spring2012_Boutin|Back to first bonus point opportunity, ECE302 Spring 2013]] | [[Bonus_point_1_ECE302_Spring2012_Boutin|Back to first bonus point opportunity, ECE302 Spring 2013]] |
Latest revision as of 10:47, 28 January 2013
Problem Involving Conditional Probability
Question:
If there are 2 white balls and 4 red balls in box 1, and 5 white balls and 3red balls in box 2. Now randomly take one ball from box 1 and put into box 2. Then take one ball from box 2.
- If under the condition that the ball being taken from box1 is red, what is the possibility that the ball that's taken from box 2 is also red?
- What's the possibility that the ball that's taken from box 2 is red?
Answer:
(1)Assume event A is the ball that's taken from box 2 is red; and event B is the ball that's taken from box 1 is red.
P(B) = 4/(2+4) = 2/3
P(A|B) =P(A_AND_B)/P(B)
= (2/3)*(4/9)/(2/3) =4/9
(2)
P(B) = 4/(2+4) = 2/3
P(NOT_B) = 1 - 2/3 = 1/3
P(A) = P(A_AND_B) + P(A_AND_NOTB)
= P(A|B)*P(B) + P(A|NOT_B)*P(NOT_B) =(4/9)*(2/3) + (1/3)*(1/3) =11/27
Comments/discussion
- Write comment here
- answer here