An executive in a company makes on an average 5 telephone calls per hour at a cost of ₹ 2 per cell. The probability that in any hour the cost of the calls exceeds a sum of ₹ 4 is

Question

An executive in a company makes on an average 5 telephone calls per hour at a cost of ₹ 2 per cell. The probability that in any hour the cost of the calls exceeds a sum of ₹ 4 is (A) (2 e4-35/2 e5) (B) (2 e5-37/2 e5) (C) 1-(37/e4) (D) 1-(18.5) e5

Tardigrade Admin · Accepted Answer

The number of telephone calls per hour is a random variable with mean

= 5

.
The required probability is given by

P (r > 2) = 1 - P (r \leq 2) = 1 - r = 0 \sum 2 \frac{e ^{- 5} \cdot 5 ^{r}}{r !}

= 1 - [\frac{5 ^{0}}{0 !} + \frac{5 ^{1}}{1 !} + \frac{5 ^{2}}{2 !}] \frac{1}{e ^{5}} = 1 - \frac{37}{2 e ^{5}} = \frac{2 e ^{5} - 37}{2 e ^{5}}

Q. An executive in a company makes on an average $5$ telephone calls per hour at a cost of $₹2$ per cell. The probability that in any hour the cost of the calls exceeds a sum of $₹4$ is

$\frac{2 e ^{4} - 35}{2 e ^{5}}$

$\frac{2 e ^{5} - 37}{2 e ^{5}}$

$1 - \frac{37}{e ^{4}}$

$1 - (18.5) e^{5}$

Q. An executive in a company makes on an average 5 telephone calls per hour at a cost of ₹2 per cell. The probability that in any hour the cost of the calls exceeds a sum of ₹4 is

2e52e4−35​

2e52e5−37​

1−e437​

1−(18.5)e5

The number of telephone calls per hour is a random variable with mean =5. The required probability is given by P(r>2)=1−P(r≤2)=1−r=0∑2​r!e−5⋅5r​ =1−[0!50​+1!51​+2!52​]e51​=1−2e537​=2e52e5−37​

Q. An executive in a company makes on an average $5$ telephone calls per hour at a cost of $₹2$ per cell. The probability that in any hour the cost of the calls exceeds a sum of $₹4$ is

$\frac{2 e ^{4} - 35}{2 e ^{5}}$

$\frac{2 e ^{5} - 37}{2 e ^{5}}$

$1 - \frac{37}{e ^{4}}$

$1 - (18.5) e^{5}$