Suppose that the number of accidents occurring on a highway each day is a Poisso

(a) To find the probability that 3 or more accidents occur today, we can use the Poisson distribution formula. The probability mass function (PMF) of a Poisson random variable with parameter $\lambda$ is given by:
$P(X=k)=\frac{e^{-\lambda }{\lambda }^k}{k!}$ where $X$ is the Poisson random variable representing the number of accidents, $\lambda$ is the parameter (in this case, $\lambda =3$), and $k$ is the number of accidents.
To find the probability that 3 or more accidents occur, we sum the probabilities for $k=3,4,5,\dots$:
$P(X\ge 3)=P(X=3)+P(X=4)+P(X=5)+\dots$
Let's calculate this probability.
$P(X\ge 3)=P(X=3)+P(X=4)+P(X=5)+\dots$
$={\sum }_{k=3}^{\infty }\frac{e^{-\lambda }{\lambda }^k}{k!}$
$={\sum }_{k=3}^{\infty }\frac{e^{-3}{3}^k}{k!}$
Calculating this sum directly can be challenging, so we can use the complementary probability approach. The complement of ''3 or more accidents occur'' is ''2 or fewer accidents occur.'' Therefore, we can calculate the complementary probability and subtract it from 1 to get the desired probability.
$P(X\ge 3)=1-P(X\le 2)$
$=1-(P(X=0)+P(X=1)+P(X=2))$
$=1-(\frac{e^{-3}{3}^0}{0!}+\frac{e^{-3}{3}^1}{1!}+\frac{e^{-3}{3}^2}{2!})$
$=1-(e^{-3}+3e^{-3}+\frac{9}{2}{e}^{-3})$
$=1-\left(\frac{11}{2}{e}^{-3}\right)$
Therefore, the probability that 3 or more accidents occur today is $\frac{11}{2}{e}^{-3}$.
(b) To find the probability that 3 or more accidents occur today, given that at least 1 accident occurs, we need to consider the conditional probability. We denote this probability as $P(X\ge 3|X\ge 1)$.
Using the definition of conditional probability, we have:
$P(X\ge 3|X\ge 1)=\frac{P(X\ge 3\cap X\ge 1)}{P(X\ge 1)}$
Since $X\ge 3$ implies $X\ge 1$, we have $P(X\ge 3\cap X\ge 1)=P(X\ge 3)$. Therefore, we can simplify the expression as:
$P(X\ge 3|X\ge 1)=\frac{P(X\ge 3)}{P(X\ge 1)}$
We already calculated $P(X\ge 3)$ in part (a), and to calculate $P(X\ge 1)$, we can use the complement again:
$P(X\ge 1)=1-P(X=0)$
$=1-\frac{e^{-3}{3}^0}{0!}$
$=1-e^{-3}$
Substituting these values into the expression for conditional probability:
$P(X\ge 3|X\ge 1)=\frac{\frac{11}{2}{e}^{-3}}{1-e^{-3}}$
Therefore, the probability that 3 or more accidents occur today, given that at least 1 accident occurs, is $\frac{\frac{11}{2}{e}^{-3}}{1-e^{-3}}$.

Step 1:
(a) To find the probability that 3 or more accidents occur today, we need to calculate the cumulative probability of the Poisson distribution up to 2 accidents and subtract it from 1. The probability mass function (PMF) of a Poisson random variable with parameter $\lambda$ is given by:
$P(X=k)=\frac{e^{-\lambda }{\lambda }^k}{k!}$ where $X$ represents the number of accidents.
Let's calculate the probability using this formula:
$P(\text{3 or more accidents})=1-P(X\le 2)$
Substituting the value of $\lambda =3$, we have:
$P(X\le 2)=P(X=0)+P(X=1)+P(X=2)$
Calculating the individual probabilities:
$P(X=0)=\frac{e^{-3}·3^0}{0!}$
$P(X=1)=\frac{e^{-3}·3^1}{1!}$
$P(X=2)=\frac{e^{-3}·3^2}{2!}$
Now, we can substitute these values and calculate the final probability:
$P(\text{3 or more accidents})=1-(\frac{e^{-3}·3^0}{0!}+\frac{e^{-3}·3^1}{1!}+\frac{e^{-3}·3^2}{2!})$
Step 2:
(b) To solve this part, we need to find the probability that at least 1 accident occurs today. This can be calculated as:
$P(\text{at least 1 accident})=1-P(X=0)$
Substituting the value of $\lambda =3$, we have:
$P(\text{at least 1 accident})=1-\frac{e^{-3}·3^0}{0!}$
Now, you can calculate the final probability using the above equation.