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Updated on: 18 Mar 2013, 06:30
Originally posted by SVaidyaraman on 17 Mar 2013, 18:53.
Last edited by SVaidyaraman on 18 Mar 2013, 06:30, edited 1 time in total.
Last edited by SVaidyaraman on 18 Mar 2013, 06:30, edited 1 time in total.
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Difficulty:
35%
(medium)
Question Stats:
66% (01:15) correct
34%
(01:37)
wrong
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The probability that B can shoot a target two out of two times is 0.25. What is the probability that the target will be missed by B immediately after such two shots?
A. 0.25
B. 0.5
C. 0.75
D. 0.4
E. 0.8
A. 0.25
B. 0.5
C. 0.75
D. 0.4
E. 0.8
18 Mar 2013, 01:46
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The probability of 2 "times" is 0,25
\(x^2=0,25\)
\(x=0,5\)
x= probability of missing/hitting the target
The probability that he will miss the next shot is 0,5.
"The streak" is not relevant.If I throw a coin the probability of getting one "side" is the same regardless of what I have obtained in the previous shots.
Correct me if I am wrong.
\(x^2=0,25\)
\(x=0,5\)
x= probability of missing/hitting the target
The probability that he will miss the next shot is 0,5.
"The streak" is not relevant.If I throw a coin the probability of getting one "side" is the same regardless of what I have obtained in the previous shots.
Correct me if I am wrong.
General Discussion
17 Mar 2013, 23:23
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Probablity of B can shoot a target 2/2 is 0.25 i.e. the probablity that he will shot both the times is 1/4.
And the probality that he will score either 1 or 0 (miss 1 or both) is 3/4.
Hence the probablity of miss the target should be 3/4 i.e. 0.75.
Answer is C.
And the probality that he will score either 1 or 0 (miss 1 or both) is 3/4.
Hence the probablity of miss the target should be 3/4 i.e. 0.75.
Answer is C.
