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18 Mar 2013, 07:03
Kudos
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
38% (01:58) correct
62%
(01:51)
wrong
based on 201
sessions
History
Date
Time
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Not Attempted Yet
The probability of shooting a target increases after a certain skill is enhanced and is equal to the new probability of NOT shooting the target. Given this fact, which of the following must be false?
A. The new probability of shooting the target is greater than 0.5
B. The original probability of shooting the target is less than 0.5
C. The original probability of NOT shooting the target and the new probability of shooting the target are the same
D. The original probability of shooting the target and that of NOT shooting the target are the same.
E. The sum of the original and the new probabilities of shooting the target is ALWAYS equal to 1.
A. The new probability of shooting the target is greater than 0.5
B. The original probability of shooting the target is less than 0.5
C. The original probability of NOT shooting the target and the new probability of shooting the target are the same
D. The original probability of shooting the target and that of NOT shooting the target are the same.
E. The sum of the original and the new probabilities of shooting the target is ALWAYS equal to 1.
20 Mar 2013, 03:12
Kudos
Bookmarks
Quote:
I think this is a 700 problem.
If I understood well the question, we have:
- initial probability of shooting target: \(P(success\;initial)\)
- initial probability of not shooting the target: \(P(not\;success\;initial)\)
- new probability of shooting target: \(P(success\;final)\)
- new probability of not shooting the target: \(P(not\;success\;final)\)
Conditions given by problem:
(1) \(P(success\;initial)=1-P(not\;success\;initial)\)
(2) \(P(success\;final)=1-P(not\;success\;final)\)
(3) \(P(success\;initial)<P(success\;final)\)
(4) \(P(success\;initial)=P(not\;success\;final)\)
Therefore:
\(P(success\;final)=1-P(success\;initial)\)
then
\(P(success\;initial)<1-P(success\;initial)\) ---> \(P(success\;initial)<0.5\)
Conclusions:
(5) \(P(success\;initial)<0.5\)
(6) \(P(success\;final)>0.5\)
(7) \(P(not\;success\;final)<0.5\)
(8) \(P(not\;success\;initial)>0.5\)
(9) \(P(success\;final)=1-P(not\;success\;final)\) ---> \(P(success\;final)=1-P(success\;initial)\) ---> \(P(success\;final)+P(success\;initial)=1\)
Analysis of different options:
A. The new probability of shooting the target is greater than 0.5: TRUE, look at (6)
B. The original probability of shooting the target is less than 0.5: TRUE, look at (5)
C. The original probability of NOT shooting the target and the new probability of shooting the target are the same: TRUE, look at (4)
D. The original probability of shooting the target and that of NOT shooting the target are the same: FALSE
E. The sum of the original and the new probabilities of shooting the target is ALWAYS equal to 1: : TRUE, look at (9)
20 Mar 2013, 08:45
Kudos
Bookmarks
In fact, the key formula here to solve the whole problem is:
\(P(A)=1-P(not\;A)\)
Then differentiate between the initial situation and the new (or final) situation:
Initial: \(P(A)=1-P(not\;A)\)
New: \(P(Z)=1-P(not\;Z)\)
And link the formulas with the conditions given by the problem.
\(P(A)=1-P(not\;A)\)
Then differentiate between the initial situation and the new (or final) situation:
Initial: \(P(A)=1-P(not\;A)\)
New: \(P(Z)=1-P(not\;Z)\)
And link the formulas with the conditions given by the problem.
