Difference between revisions of "GATE2013 q2"

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)

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

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

@@ Line 1: / Line 1: @@
-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?
+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) $8/(2e^{3})$
@@ Line 5: / Line 5: @@
 (B) $9/(2e^{3})$
-(C) $17/(2e^{3})$
+'''(C) $17/(2e^{3})$'''
 (D) $26/(2e^{3})$
+==={{Template:Author|Arjun Suresh|{{arjunweb}} }}===
+Poisson Probability Density Function (with mean $\lambda$) = $\lambda^{k} / (e^{\lambda}k!)$,
-<\disqus>
+We have to sum the probability density function for $k = 0,1$ and $2$ and $\lambda$ = 3 (thus finding the cumulative mass function)
+=$(1/e^3) + (3/e^3) + (9/2e^3)$
+=$17/(2e^{3})$
+{{Template:FBD}}
+[[Category: GATE2013]]
+[[Category: Probability questions from GATE]]