Difference between revisions of "GATE2013 q2"

 
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Suppose <math>p</math> is the number of cars per minute passing through a certain road junction between 5 PM and 6 PM, and <math>p</math> has a Poisson distribution with mean 3. What is the probability of observing fewer than 3 cars during any given minute in this interval?
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Suppose <math>p</math> is the number of cars per minute passing through a certain road junction between 5 PM and 6 PM, and <math>p</math> has a Poisson distribution with mean $3$. What is the probability of observing fewer than 3 cars during any given minute in this interval?
  
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(A) $8/(2e^{3})$
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<input type = "radio"  name="2" value="A" onclick="handleOption(this);">          (A) $8/(2e^{3})$
 
  
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(B) $9/(2e^{3})$
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<input type = "radio"  name="2" value="B" onclick="handleOption(this);">(B) $9/(2e^{3})$
 
  
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'''(C) $17/(2e^{3})$'''
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<input type = "radio"  name="2" value="C" onclick="handleOption(this);">                                    (C) $17/(2e^{3})$
 
  
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(D) $26/(2e^{3})$
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<input type = "radio"  name="2" value="D" onclick="handleOption(this);">                                    (D) $26/(2e^{3})$
 
  
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==={{Template:Author|Arjun Suresh|{{arjunweb}} }}===
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Poisson Probability Density Function (with mean $\lambda$) = $\lambda^{k} / (e^{\lambda}k!)$,
  
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We have to sum the probability density function for $k = 0,1$ and $2$ and $\lambda$ = 3 (thus finding the cumulative mass function)
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<input type = "radio"  name="2" value="E" checked = "checked" onclick="handleNooption(this);"> Unmarked
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=$(1/e^3) + (3/e^3) + (9/2e^3)$
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=$17/(2e^{3})$
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<button type="button" name="2" onclick="handleMarkme(this);">Check later </button>
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[[Category: GATE2013]]
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[[Category: Probability questions from GATE]]

Latest revision as of 11:53, 15 July 2014

Suppose <math>p</math> is the number of cars per minute passing through a certain road junction between 5 PM and 6 PM, and <math>p</math> has a Poisson distribution with mean $3$. What is the probability of observing fewer than 3 cars during any given minute in this interval?

(A) $8/(2e^{3})$

(B) $9/(2e^{3})$

(C) $17/(2e^{3})$

(D) $26/(2e^{3})$

Solution by Arjun Suresh

Poisson Probability Density Function (with mean $\lambda$) = $\lambda^{k} / (e^{\lambda}k!)$,

We have to sum the probability density function for $k = 0,1$ and $2$ and $\lambda$ = 3 (thus finding the cumulative mass function)

=$(1/e^3) + (3/e^3) + (9/2e^3)$

=$17/(2e^{3})$




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Suppose <math>p</math> is the number of cars per minute passing through a certain road junction between 5 PM and 6 PM, and <math>p</math> has a Poisson distribution with mean 3. What is the probability of observing fewer than 3 cars during any given minute in this interval?

</div>

<input type = "radio" name="2" value="A" onclick="handleOption(this);"> (A) $8/(2e^{3})$

<input type = "radio" name="2" value="B" onclick="handleOption(this);">(B) $9/(2e^{3})$

<input type = "radio" name="2" value="C" onclick="handleOption(this);"> (C) $17/(2e^{3})$

<input type = "radio" name="2" value="D" onclick="handleOption(this);"> (D) $26/(2e^{3})$


<input type = "radio" name="2" value="E" checked = "checked" onclick="handleNooption(this);"> Unmarked

<button type="button" name="2" onclick="handleMarkme(this);">Check later </button>

</div>