Probability Practice Exam — Free

Define the events $D_1, D_2, D_3$ (the parcel came from that depot) and $L$ (the parcel is late). The given data are the priors $P(D_1) = 0.5$, $P(D_2) = 0.3$, $P(D_3) = 0.2$ and the conditional late rates $P(L \mid D_1) = 0.01$, $P(L \mid D_2) = 0.02$, $P(L \mid D_3) = 0.05$. We want the reverse conditional probability $P(D_3 \mid L)$, which is what Bayes' theorem provides.

First find $P(L)$ by the law of total probability, since the three depots partition the parcels:

Then apply Bayes' theorem:

So although Depot 3 ships only a fifth of the parcels, it is responsible for nearly half of the late ones, because its late rate is much higher than the others'.

The tempting wrong answer is 0.625, which is $\dfrac{0.05}{0.01 + 0.02 + 0.05}$: it normalizes the late rates without weighting them by how many parcels each depot ships, so it ignores the priors. The answer 0.010 is the joint probability $P(L \cap D_3)$, not the conditional one, and 0.200 is the prior $P(D_3)$ before the late arrival is taken into account.

$P(D_3 \mid L) = \dfrac{10}{21} \approx 0.476$ (option C).

Let $X$ be the number of flaws in the 5 km length. For a Poisson process the count in a fixed length is Poisson distributed, with mean equal to the rate times the length, because flaws occur independently and uniformly along the cable:

So $X \sim \text{Poisson}(2)$ and

Substitute $k = 2$ and $\lambda = 2$:

Numerically, $e^{-2} \approx 0.135335$, so $P(X = 2) \approx 0.2707$.

A common error is to use $\lambda = 0.4$ (the rate per kilometre) without scaling to the 5 km length; that would give $P(X = 2) = \dfrac{e^{-0.4}(0.4)^2}{2} \approx 0.0536$, the probability of 2 flaws in only 1 km. Another is to forget the $k!$ in the denominator, which gives $4e^{-2} \approx 0.541$.

$P(X = 2) = 2e^{-2} \approx 0.2707$

(a) The joint probabilities are $p(1,1) = \tfrac{2}{21}$, $p(1,2) = \tfrac{3}{21}$, $p(1,3) = \tfrac{4}{21}$, $p(2,1) = \tfrac{3}{21}$, $p(2,2) = \tfrac{4}{21}$, $p(2,3) = \tfrac{5}{21}$; they sum to $\tfrac{21}{21} = 1$, as a pmf must. Marginals are obtained by summing over the other variable:

(b) Expectations are probability-weighted sums:

For $E[XY]$, sum $xy\,p(x,y)$ over all six pairs:

(c) Use the computational formula $\operatorname{Cov}(X,Y) = E[XY] - E[X]E[Y]$:

Independence requires $p(x,y) = p_X(x)\,p_Y(y)$ for every pair. It fails already at $(1,1)$: $p_X(1)\,p_Y(1) = \dfrac{9}{21}\cdot\dfrac{5}{21} = \dfrac{45}{441} \approx 0.102$, whereas $p(1,1) = \dfrac{2}{21} \approx 0.095$. So $X$ and $Y$ are not independent. (A nonzero covariance also rules out independence on its own, since independent variables always have covariance 0; the converse is false, which is why the pmf comparison is the decisive test.)

$E[X] = \dfrac{11}{7}$, $E[Y] = \dfrac{46}{21}$, $E[XY] = \dfrac{24}{7}$, $\operatorname{Cov}(X,Y) = -\dfrac{2}{147} \approx -0.0136$; $X$ and $Y$ are not independent.