Answer to Question #206375 in Statistics and Probability for Chilu

Question

I. A sales manager receives 6 calls on average between 9:30 a.m. and 10:30 a.m. on a weekday. Find the probability that: a) he will receive 2 or more calls between 9:30 a.m. and 10:30 a.m. on a certain weekday. b) he wiI1 receive exactly 2 calls between 9:30 a.m. and 9:40 a.m. on a certain weekday. c) during a 5 day working week, there will be exactly 3 days on which he will receive no calls between 9:30 a.m. and 9:40 a.m. II. Suppose an ordinary six-sided die is rolled repeatedly and the outcome is noted on each roll. What is the probability that the: a) third 6 occurs on the seventh roll? b) number of rolls until the first 6 occurs is at most 10? 

Answer 1

(1) here,

$\lambda=6$

(a) $P(X\ge 2)=1-[P(X=0)+P(X=1)]$

$=1-[\dfrac{e^{-\lambda} \lambda^0}{0!}+\dfrac{e^{-\lambda} \lambda^1}{1!}] \\[9pt] =1-[\dfrac{e^{-6} 6^0}{0!}+\dfrac{e^{-6} 6^1}{1!}] \\[9pt] =1-0.01735=0.98234$

(b) $P(X=2)=\dfrac{e^{-\lambda } \lambda^2}{2!}=\dfrac{e^{-6}. 6^2}{2!}=0.0446$

(c) In 3 days Probability that exactly 3 days on which he do not recieve any call-

$=3\times \dfrac{e^{-\lambda} \lambda^0}{0!}=3\dfrac{e^{-6} 6^0}{0!}=0.007436$

(2)

(a) Probability that 3 six occur on the 7 roll-

=$^6C_2(\dfrac{1}{6})^2(\dfrac{5}{6})^4\times \dfrac{1}{6}=0.0334$

(b) Probability that the number of rolls until the first 6 occurs is at most 10-

$=1-^{10}C_{0} (\dfrac{1}{6})^0(\dfrac{5}{6})^{(10-0)}\times \dfrac{1}{6}=1-0.026917=0.97308$