Answer
Verified
392.4k+ views
Hint: We can solve the given question using Poisson distribution formula \[{\text{P}}\left( {{\text{X = r}}} \right) = \dfrac{{{{\text{e}}^{ - {\text{m}}}}{{\text{m}}^{\text{r}}}}}{{{\text{r!}}}}\] where ‘m’ is the expected number of occurrences or the average number of events in a given time interval and ‘X’ is the number of events observed over a given time period .Put the given values in the formula and simplify it. Here, e is the base of natural logarithm (also called Euler’s number)
Step-by-Step solution:
Given, an average number of phone calls during $10$-minute time intervals(m)=$5$.We know that the number of phone calls at the telephone enquiry system follows the Poisson distribution. We have to find the probability that there is at the most one phone call during a $10$-minute time intervals. So here X≤$1$.Now, according to Poisson distribution, \[{\text{P}}\left( {{\text{X = r}}} \right) = \dfrac{{{{\text{e}}^{ - {\text{m}}}}{{\text{m}}^{\text{r}}}}}{{{\text{r!}}}}\] where, ‘m’ is the expected number of occurrences or the average number of events in a given time interval and ‘X’ is the number of events observed over a given time period. Here, e is the base of the natural logarithm (also called Euler’s number). Now, for there to be at the most one phone call to happen during the given interval, the value of X must be equal to or greater than one.
∴${\text{P}}\left( {{\text{X}} \leqslant {\text{1}}} \right) = {\text{P}}\left( {{\text{X = 0}}} \right) + {\text{P}}\left( {{\text{X}} = 1} \right)$
On putting the values in the formula we get, \[{\text{P}}\left( {{\text{X}} \leqslant {\text{1}}} \right) = \dfrac{{{{\text{e}}^{ - {\text{m}}}}{{\text{m}}^0}}}{{{\text{0!}}}} + \dfrac{{{{\text{e}}^{ - {\text{m}}}}{{\text{m}}^1}}}{{{\text{1!}}}} = {{\text{e}}^{ - 5}} + (5 \times {{\text{e}}^{ - 5}})\]
Om simplifying it, we get,
$ \Rightarrow {{\text{e}}^{ - 5}}\left( {1 + 5} \right) = \dfrac{6}{{{{\text{e}}^5}}}$
Hence the answer is ‘D’.
Note: Poisson distribution is used to predict the probability of certain events from happening when you know how often the event has occurred. The condition for Poisson distribution is-
1. An event can occur any number of times during a time period.
2. Events occur independently. Example-for the number of phone calls an office would receive, there is no reason to expect a caller to affect the chances of another person calling.
3. The rate of occurrence is constant and not based on time.
4. The probability of an event occurring is proportional to the length of the time period.
Step-by-Step solution:
Given, an average number of phone calls during $10$-minute time intervals(m)=$5$.We know that the number of phone calls at the telephone enquiry system follows the Poisson distribution. We have to find the probability that there is at the most one phone call during a $10$-minute time intervals. So here X≤$1$.Now, according to Poisson distribution, \[{\text{P}}\left( {{\text{X = r}}} \right) = \dfrac{{{{\text{e}}^{ - {\text{m}}}}{{\text{m}}^{\text{r}}}}}{{{\text{r!}}}}\] where, ‘m’ is the expected number of occurrences or the average number of events in a given time interval and ‘X’ is the number of events observed over a given time period. Here, e is the base of the natural logarithm (also called Euler’s number). Now, for there to be at the most one phone call to happen during the given interval, the value of X must be equal to or greater than one.
∴${\text{P}}\left( {{\text{X}} \leqslant {\text{1}}} \right) = {\text{P}}\left( {{\text{X = 0}}} \right) + {\text{P}}\left( {{\text{X}} = 1} \right)$
On putting the values in the formula we get, \[{\text{P}}\left( {{\text{X}} \leqslant {\text{1}}} \right) = \dfrac{{{{\text{e}}^{ - {\text{m}}}}{{\text{m}}^0}}}{{{\text{0!}}}} + \dfrac{{{{\text{e}}^{ - {\text{m}}}}{{\text{m}}^1}}}{{{\text{1!}}}} = {{\text{e}}^{ - 5}} + (5 \times {{\text{e}}^{ - 5}})\]
Om simplifying it, we get,
$ \Rightarrow {{\text{e}}^{ - 5}}\left( {1 + 5} \right) = \dfrac{6}{{{{\text{e}}^5}}}$
Hence the answer is ‘D’.
Note: Poisson distribution is used to predict the probability of certain events from happening when you know how often the event has occurred. The condition for Poisson distribution is-
1. An event can occur any number of times during a time period.
2. Events occur independently. Example-for the number of phone calls an office would receive, there is no reason to expect a caller to affect the chances of another person calling.
3. The rate of occurrence is constant and not based on time.
4. The probability of an event occurring is proportional to the length of the time period.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
Using the following information to help you answer class 12 chemistry CBSE
Trending doubts
State the differences between manure and fertilize class 8 biology CBSE
Why are xylem and phloem called complex tissues aBoth class 11 biology CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
What would happen if plasma membrane ruptures or breaks class 11 biology CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
What precautions do you take while observing the nucleus class 11 biology CBSE
What would happen to the life of a cell if there was class 11 biology CBSE
Change the following sentences into negative and interrogative class 10 english CBSE