Question
Mathematics Question on Continuity and differentiability
It is known from past experience that in a certain plant there are on the average 4 industrial accidents per month. Find the probability that there will be less than 4 accidents in a given month. Given: (e−4=0.0183)
A
(A) 0.423
B
(B) 0.433
C
(C) 0.443
D
(D) 0.453
Answer
(B) 0.433
Explanation
Solution
Explanation:
Given:m=4The probability distribution is given by:p(x)=(e−m×mx)x!Calculation:According to the formula,⇒p(x)=(e−4×4x)x!Now, the probability that in a given month there will be less than 4 acc idents=p(0)+p(1)+p(2)+p(3)=∑x=0x=3e−44xx!=e−4[(1+4+8)+323]=e−4[(13)+323]=0.0183×713=0.0183×23.666=0.433∴ The probability that in a given month there will be less than 4 accidents is 0.433.Hence, the correct option is (B).