0 votes 0 votes Shankar Kakde asked Oct 1, 2018 Shankar Kakde 421 views answer comment Share Follow See 1 comment See all 1 1 comment reply arvin commented Oct 2, 2018 reply Follow Share 0.224 0 votes 0 votes Please log in or register to add a comment.
1 votes 1 votes $\lambda = 3$ $P(k) = e^{- \lambda} \frac{\lambda ^k}{ k!}$ $P(2) =$$\large e^{- 3} \times \frac{3 ^2}{ 2!}$ $P(2) = 0.22404$ Mk Utkarsh answered Oct 2, 2018 • edited Oct 3, 2018 by Mk Utkarsh Mk Utkarsh comment Share Follow See all 2 Comments See all 2 2 Comments reply Mayankprakash commented Nov 12, 2018 reply Follow Share @utkarsh How do we get know lambda and k from the question? 0 votes 0 votes Mk Utkarsh commented Nov 12, 2018 reply Follow Share 180 calls per hour so 3 calls per minute 0 votes 0 votes Please log in or register to add a comment.
