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10 Mar 2018, 05:18
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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25% (01:15) correct
75%
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wrong
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The probability that at least one company is going to announce a takeover in the next hour is 84%. What is the probability that at least one announces in the next 30 minutes? ( assuming that the distribution of announcement time is independently uniformly distributed )
A) 40%
B) 42%
C) 56%
D) 60%
E) 84%
Source : Other - Brilliant.
A) 40%
B) 42%
C) 56%
D) 60%
E) 84%
Source : Other - Brilliant.
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Transponster
10 Mar 2018, 06:37
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Since distribution of announcement time is independently uniformly distributed
The number of distributions per hour has mean λ and is Poisson
p(X≥1)=1−p(X=0)=1−e^(−λ)=0.84
⟹λ=−ln0.16
So, the probability of announcement in 30 minutes is
1−e(-30λ/60)=1−e^ln0.4=0.6
There fore, answer is 60%
The number of distributions per hour has mean λ and is Poisson
p(X≥1)=1−p(X=0)=1−e^(−λ)=0.84
⟹λ=−ln0.16
So, the probability of announcement in 30 minutes is
1−e(-30λ/60)=1−e^ln0.4=0.6
There fore, answer is 60%
10 Mar 2018, 06:55
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swetakumari
I am not sure I completely understand the math being used in there. Can you elaborate the steps used a bit further?
Thank you for posting!
