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16 Apr 2018, 05:57
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
14% (02:24) correct
86%
(01:13)
wrong
based on 7
sessions
History
Date
Time
Result
Not Attempted Yet
The probability of success in a field event is the ninth part of the probability of failure. Successive and independent trials are held until success occurs for the first time. In these circumstances, the expected number of failures that should occur until the first success occurs is equal to:
(A) 2
(B) 3
(C) 6
(D) 9
(E) 10
(A) 2
(B) 3
(C) 6
(D) 9
(E) 10
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16 Apr 2018, 06:13
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CursoFDX
From the question stem, we understand that, probability of success P(s) is 1/9 of probability of failure P(f).
P(s) = 1/9 P(f).
Now analyzing from the point of view of options, if the number of trials before first success is 2 then P(s) =1/2 and P(f) = 1/2 (Wrong, Not what question stem mentions)
From option B, P(s) = 1/3, P(f) = 2/3, (Wrong, question stem needs P(s) = 1/9P(f)
from option C, P(s) = 1/6, P(f) = 5/6. (Wrong)
From option D , P(s) = 1/9, P(f) = 8/9 (Wrong)
From option E, P(s) = 1/10, P(f) = 9/10 => P(s)/P(f) = 1/9 => P(s)=1/9 P(f) (Right Answer)
Hence option E is right answer
16 Apr 2018, 06:23
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CursoFDX
Let Failure = x, so Success= x/9. Since S+F=1 --> x + x/9 = 1 --> F=9/10 and S=1/10
{9F}+{1S}=10. I got max 9 Failures before 1 Success, why 10 F (E)? The Q asks how many F not trials why 10 and not 9???
Thanks.
