Practice Problems (Introduction to Probability Theory)


A few practice problems related to Probability theory are given here. These problems are adapted from the textbook: Probability and Random Processes by Scott Miller 2nd Edition.

These problems cover the following topics:

  1. Experiments, Sample spaces, events,
  2. Axioms of Probability, Assigning Probability,
  3. Joint and Conditional Probability, Independence,
  4. Baye’s Theorem, Total Law of Probability,
  5. Discrete Random variables.

Problem-1: A roulette wheel consists of 38 numbers (18 are red, 18 are black, and 2 are green). Assume that with each spin of the wheel, each number is equally likely to appear.

(a) What is the probability of a gambler winning if he bets on a red number showing up?
(b) Suppose the gambler keeps betting on red until he finally wins. Let N be the number of times he plays/bets. Specify the probability mass function of the random variable N. That is, find \(P_N(k) = P(N=k)\).
(c) Now, suppose the gambler keeps betting on red until he wins twice. Let M be the number of times he plays/bets. Specify the probability mass function of the random variable M . That is, find \(P_M(k) = P(M=k)\).

Solution: (a) Pr(red) = 18/38 \(\approx\) 0.4737

(b) Let N be the number of plays/bets and the player bets on red until he wins.

Pr(N=1) = \(18/38\)

Pr(N=2) = \((20/38)(18/38)\)

Pr(N=3) = \((20/38)^{2}(18/38)\)

Pr(N=k) = \((20/38)^{k-1}(18/38)\)

(c) Now let M be the number of plays/bets and the gambler bets on red until he wins twice.

Pr(M=2) = \((18/38)^{2}\)

Pr(M=3) = \((3 -1) (20/38)(18/38)^{2}\)

Pr(M=4) = \((4-1) (20/38)^{2}(18/38)^{2}\)

Pr(M=k) = \((k-1) (20/38)^{k-2}(18/38)^{2}\)

Problem-2: Highway A and Highway B merge to form Highway C as shown in the figure. Engineers have studied the traffic patterns on the two merging roads and found that the number of cars per minute that travel each road can be well modelled as Poisson random variables as described below:

highway_problem_probability

Highway A: N=# cars per minute, \(P(N=n) = \frac{(\lambda_A)^n e^{-\lambda_A}}{n!} \;,\;\; n=0,1,2,3,\ldots\)

Highway B: M=# cars per minute, \(P(M=m) = \frac{(\lambda_B)^m e^{-\lambda_B}}{m!} \;,\;\; m=0,1,2,3,\ldots\)

Let K = M+N be the number of cars per minute on Highway C. Find the PMF of K. Is K also a Poisson random variable or does it follow some other distribution? You may assume that in any interval of time, the number of cars on Highway A and the number of cars on Highway B are independent of each other.

Solution: \( Pr(K=k)=Pr\Big(\bigcup_{m=0}^{k}[\{M=m\}\cap\{N=k-m\}]\Big)\)
\(=\sum_{k+1}^{k}Pr(M=m,N=k-m)\)
\(=\sum_{m=0}^{k}Pr(M=m)Pr(N=k-m)\)
\(=\sum_{m=0}^{k}\frac{(\lambda_{B})^{m}e^{-\lambda_{B}}}{m!}\frac{(\lambda_{A})^{k-m}e^{-\lambda_{A}}}{(k-m)!}\)
\(=e^{-\lambda_{B}}e^{-\lambda_{A}}(\lambda_{A})^{k}\sum_{m=0}^{k}\frac{(\lambda_{B})^{m}(\lambda_{A})^{-m}}{(m!)(k-m)!}\)
Multiplying and dividing by k!

\(=\frac{e^{-(\lambda_{B}+\lambda_{A})}(\lambda_{A})^{k}}{k!}\sum_{m=0}^{k}\frac{k!}{m!(k-m)!}{\bigg(\frac{\lambda_{B}}{\lambda_{A}}\bigg)}^{m}\)
\(=\frac{e^{-(\lambda_{B}+\lambda_{A})}(\lambda_{A})^{k}}{k!}\sum_{m=0}^{k}{k \choose m}{\bigg(\frac{\lambda_{B}}{\lambda_{A}}\bigg)}^{m}\)
\(=\frac{e^{-(\lambda_{B}+\lambda_{A})}(\lambda_{A})^{k}}{k!}{\bigg(1+\frac{\lambda_{B}}{\lambda_{A}}\bigg)}^{k}\)
\(=\frac{e^{-(\lambda_{B}+\lambda_{A})}(\lambda_{A}+\lambda_{B})^{k}}{k!}\)
This follows a Poisson random variable.

Problem-3: In the game of RISK, two players compete in a game of dice rolling for conquest of the world. One player is on “offense” while the other is on “defense.” For this problem, the player on offense is allowed to roll multiple dice while the player on defense rolls a single die. Whoever rolls the higher number wins (i.e., the highest number rolled by the offense is compared with the number rolled by the defense). In case of a tie, the defense is declared the winner. The loser must remove one army from the board. Find the probability of the offense winning and the probability of the defense winning in each of the following scenarios:
(a) Both players roll only a single die.
(b) Offense rolls two dice while defense rolls one die.
(c) Offense rolls three dice while defense rolls one die.

Solution: The highest number rolled wins. If there is a tie, the defense is the winner.
Let D stand for defense and O for offense.

(a) Both players roll a single die:

Pr(O wins) = Pr(O wins | O rolls 1) Pr(O rolls 1) + Pr(O wins | O rolls 2) Pr(O rolls 2) + Pr(O wins | O rolls 3) Pr(O rolls 3) + Pr(O wins | O rolls 4) Pr(O rolls 4) + Pr(O wins | O rolls 5) Pr(O rolls 5)+ Pr(O wins | O rolls 6) Pr(O rolls 6)

\(\text{Pr(O wins)}=\frac{1}{6}\bigg(\frac{0}{6}+\frac{1}{6}+\frac{2}{6}+\frac{3}{6}+\frac{4}{6}+\frac{5}{6}\bigg)=\frac{15}{36}\approx0.4167.\)
Therefore Pr(D wins) = 1 – Pr(O wins) \(\approx\) 0.5833.

(b) Offense rolls two dice while defense rolls one die: With three dice being rolled, there will be 6 \(\times\) 6 \(\times\) 6= 216 different possible outcomes. The offense will win if his/her highest roll is higher than the defense’s roll.

Pr(O wins) = Pr(O highest rolls is 1) Pr(O wins | O highest roll is 1) + …+ Pr(O highest rolls is 6) Pr(O wins | O highest roll is 6)
\(\text{Pr(O wins)}=\frac{1}{36}*0+\frac{3}{36}*\frac{1}{6}+\frac{5}{36}*\frac{2}{6}+\frac{7}{36}*\frac{3}{6}+\frac{9}{36}*\frac{4}{6}+\frac{11}{36}*\frac{5}{6}=\frac{125}{216}\approx0.5787.\)
Therefore Pr(D wins) \(\approx\) 0.4213.

(c) Offense rolls three dice while defense rolls one die:

As in Part (b): Pr(O wins) = Pr(O highest rolls is 1) Pr(O wins | O highest roll is 1) + …+ Pr(O highest rolls is 6) Pr(O wins | O highest roll is 6)
\(\text{Pr(O wins)}=\frac{1}{216}*0+\frac{7}{216}*\frac{1}{6}+\frac{19}{216}*\frac{2}{6}+\frac{37}{216}*\frac{3}{6}+\frac{61}{216}*\frac{4}{6}+\frac{91}{216}*\frac{5}{6}=\frac{855}{1296}\approx0.6597.\)
Therefore Pr(D wins) \(\approx\) 0.3403.

Problem-4: A certain light bulb manufacturer makes two types of bulbs, a low-cost short-life (Stype) bulb and a more expensive long-life (L-type) bulb. The two types of bulbs look identical and so the company must be sure to carefully label the boxes of bulbs. A box of bulbs is found on the floor of the manufacturing plant that (you guessed it) has not been labelled. In order to determine which types of bulbs are in the box, a promising young engineer suggested that they take one bulb from the box and run it until it burns out. After observing how long the bulb remains lit, they should be able to make a good guess as to which type of bulbs are in the box. It is known that the length of time (in hours), X, that a bulb lasts can be described by a geometric random variable

\(P_X(k) = (1-a)a^k \; , \;\; k=0,1,2,\ldots\)

The parameter a that appears in the above expression is a=0.99 for the S-type bulbs and a=0.999 for the L-type bulbs. It is known that of all the light bulbs the company manufactures 75% are S-type and 25% are L-type. Hence, before the experiment is run, the box in question has a 75% chance of being S-type and 25% chance of being L-type.

(a) If, after running the proposed experiment, it is observed that the bulb burned out after 200 hours, which type of bulb is most likely in the unmarked box? Mathematically justify your answer.
(b) What is the probability that your decision in part (a) turns out to be wrong? That is, if you decided that the box most likely contained L-type bulbs, what is the probability that the box actually contains S-type bulbs (or if you decided the box most likely contained S-type bulbs, what is the probability that the box actually contains L-type bulbs)?

Solution:  (a) Given Pr(bulb lasts k hours) = \((1-a)a^{k}\)

So if the bulb is S-type: Pr(bulb lasts exactly 200 hours) = \((0.01)(0.99^{200}) \approx  0.00134 \)

If the bulb is L-type: Pr(bulb lasts exactly 200 hours) = \((0.001)(0.999^{200}) \approx 0.000819\)

Using Bayes Rule: \(
\text{Pr(S-type | 200 hours)}=\frac{\text{Pr(200 hours | S-type) Pr(S-type) }}{\text{Pr(200 hours | S-type) Pr(S-type) + Pr(200 hours | L-type) Pr(L-type) }}\)
\( =\frac{\Big(\frac{3}{4}\Big)\Big(0.01\Big)\Big(0.99{^{2}}{^{0}}{^{0}}\Big)}{\Big(\frac{3}{4}\Big)\Big(0.01\Big)\Big(0.99{^{2}}{^{0}}{^{0}}\Big)\Big(\frac{1}{4}\Big)\Big(0.001\Big)\Big(0.999{^{2}}{^{0}}{^{0}}\Big)} \) \(\;\; \approx 0.8308. \)
Therefore, the unmarked box is most likely the S-type.

(b) Pr(error) = Pr(L-type | 200 hours) = 1 – Pr(S-type | 200 hours) = 0.1692.