Answer: a) 0.0000002, b) 15, c) 0.99996.
Step-by-step explanation:
Since we have given that
n = 500
p = 3% = 0.03
q = 1-0.03 = 0.97
So, we will use "Binomial distribution".
(a) What is the probability that none fail during the warranty period?
[tex]P(X=0)=^{500}C_0(0.03)^0(0.97)^{500}=0.0000002[/tex]
(b) What is the expected number of failures during the warranty period?
[tex]E(X)=np=500\times 0.03=15[/tex]
(c) What is the probability that more than 2 units fail during the warranty period?
[tex]P(X>2)=1-\sum _{x=0}^2P(X=x)=0.99996[/tex]
Hence, a) 0.0000002, b) 15, c) 0.99996.
