Assume that, during each second, a Dartmouth switchboard receives one call with probability .01 and no calls with probability .99. Use the Poisson approximation to estimate the probability that the operator will miss at most one call if she takes a 5-minute coffee break.

• 0.85
• 0.875
• 0.9
• 0.925
• 0.95

References

Doyle, Peter G. (2006). "Grinstead and Snell's Introduction to Probability" (PDF). Retrieved June 6, 2024.

Solution: E

The number of calls received after [math]t[/math] seconds equals [math]N_t[/math] has a Poisson distribution with mean [math]\lambda t [/math] where [math]\lambda [/math] is the expected number of calls after 1 second, namely 0.01. Hence the probability that we have at most one call during a 5 minute break equals approximately

[[math]]1-\exp(-0.01 * 60*5) = 0.95[[/math]]