Question

I can't figure out the formula to use for the following word problem: A fire department that on any particular day there is a 20% chance that it will have to respond to a call. What is the probability that during any 3 day span it will have to respond to at LEAST one call

Answer 1

Surely sounds like a problem in binomial probability. Number of trials (n) is 3; probability of "success" (i. e., of having to respond to a call) (p) is 0.2, and number of "successes" (x) is zero (0). Using any one of the following (table of binomial probability, the binompdf function on a TI-84 calculator, or the formula for P(x), calculate the probability of having to respond to 0 fires as binomcdf(3,.2,0). Subtracting this result from 1 gives the probability of having to respond to 1 or more fires (i. e., "at least one fire").

Answer 2

So the answer is 49%? If rounded to the nearest whole percent.

Answer 3

That looks right:
$$
\large{
\text{Probability of responding to at least 1 call}\\
=\text{1 - Probability of responding to no calls}\\
=1-\binom{3}{0}0.2^0\times 0.8^3\\
=0.488
}
$$

Answer 4

Thanks!!!

Answer 5

np