Probability is a number from 0 to 1 that measures how likely an event is. P = 0 means impossible (rolling a 7 on a standard die). P = 1 means certain (rolling any number from 1–6). P = 0.5 is a 50-50 chance (like a fair coin flip). For equally likely outcomes, count how many outcomes satisfy your event and divide by the total. The complement rule ($P(\text{not }A) = 1 - P(A)$) is often easier — instead of counting what you want, count what you don't want and subtract. The addition rule adds probabilities of two events but subtracts the overlap to avoid double-counting.
A box contains 5 defective and 45 good bolts. If one bolt is selected, find the probability that it is not defective.
$$P(\text{not defective})=1-\frac{5}{50}=0.90$$
Final answer: 0.90 or 90%.
Two Dice: Sum of 7 or 11
Two fair dice are rolled. Find the probability the sum is 7 or 11.
Total outcomes = 36. Ways to get sum 7: (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) = 6 ways. Ways to get sum 11: (5,6),(6,5) = 2 ways. Events are mutually exclusive (no overlap).
$$P=\frac{6+2}{36}=\frac{8}{36}=\frac{2}{9}$$
Final answer: $2/9 \approx 0.222$.
Mutually Exclusive Events
In a single roll of a die, find the probability of getting a 2 or a 5.
These events are mutually exclusive (both cannot happen at once), so $P(A \cap B) = 0$.
$$P(\text{at least 2 H})=\frac{4}{8}=\frac{1}{2}$$
Final answer: $1/2$.
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Exam Generator Problems
Additional board-style practice items for this topic.
Question Bank: q694
MSTE - Statistics and Probability / Probability / Engr. Janclyde Espinosa (Clidez)
If 23 people are in a room, the chances are better than even that at least two people share the same birthday.
How many people must be present in order to provide at least an even chance that two or more were born on the same day of the week?
4
14
22
6
For days of the week, no shared birthday weekday among $n$ people has probability: $\frac{7\cdot6\cdot5\cdots(7-n+1)}{7^n}$ For $n=3$, probability of a match is $1-\frac{7\cdot6\cdot5}{7^3}=0.388$. For $n=4$: $1-\frac{7\cdot6\cdot5\cdot4}{7^4}=0.650$ The first at least even chance is: $\boxed{4}$
Question Bank: q709
MSTE - Statistics and Probability / Probability / Engr. Janclyde Espinosa (Clidez)
Assume that a single depth charge has a probability 1/2 of sinking a submarine, 1/4 of damaging it, and 1/4 of missing. Assume also that two damaging explosions sink the sub.
What is the probability that 4 depth charges will sink the sub?
251/256
51/256
25/256
125/256
A submarine is not sunk only if fewer than two damaging events occur and no direct sink occurs. For each charge: sink $=1/2$, damage $=1/4$, miss $=1/4$. Not sunk after 4 charges means all non-sink outcomes with 0 or 1 damage: $P(0D,4M)=(1/4)^4$ $P(1D,3M)=\binom{4}{1}(1/4)(1/4)^3$ $P(\text{not sunk})=5/256$ $P(\text{sunk})=1-5/256$ $\boxed{251/256}$
Question Bank: q714
MSTE - Statistics and Probability / Fundamental Principles of Counting / Engr. Janclyde Espinosa (Clidez)
There are 3 red chips and 2 blue chips. When arranged in a row, they form a certain color pattern, for example RBRRB.
How many color patterns are possible?
10
12
24
60
Arrange 3 red and 2 blue chips. Number of distinct color patterns is: $\frac{5!}{3!2!}=10$ $\boxed{10}$
Question Bank: q718
MSTE - Statistics and Probability / Probability / Engr. Janclyde Espinosa (Clidez)
Kate and David each have €10. Together they flip a coin 5 times.
Every time the coin lands on heads, Kate gives David €1.
Every time the coin lands on tails, David gives Kate €1.
After the coin is flipped 5 times, what is the probability that Kate has more than €10 but less than €15?
15/32
5/16
1/2
21/32
If $H$ heads occur, Kate gives $H$ euros and receives $5-H$ euros, so her final amount is $$10-H+(5-H)=15-2H$$ The condition $10<15-2H<15$ gives $H=1$ or $H=2$. Therefore $$P=\frac{\binom51+\binom52}{2^5}=\frac{5+10}{32}=\frac{15}{32}$$ The probability is $15/32$.
Question Bank: q721
MSTE - Statistics and Probability / Fundamental Principles of Counting / Engr. Janclyde Espinosa (Clidez)
Paula has 3 movies that she can watch during the weekend:
1 Action,
1 Comedy,
1 Drama.
However, she needs to watch the Drama 3 times.
Assuming Paula has time for 5 movies and intends to watch all of them, in how many ways can she do so?
20
6
24
60
The five movie slots consist of A, C, and three D's. Distinct arrangements: $\frac{5!}{3!}=20$ $\boxed{20}$
Question Bank: q725
MSTE - Statistics and Probability / Statistics / Engr. Janclyde Espinosa (Clidez)
A spring of natural length 10 cm is such that a force of 6 kN stretches it 2 cm.
Find the work (in N·m) necessary to stretch the spring from a length of 14 cm to 18 cm.
720
680
240
350
Natural length is 10 cm. A force of 6000 N stretches it 2 cm = 0.02 m, so: $k=6000/0.02=300000\text{ N/m}$ Stretching from 14 cm to 18 cm means extension from 0.04 m to 0.08 m. Work: $W=\frac{1}{2}k(x_2^2-x_1^2)$ $W=\frac{1}{2}(300000)(0.08^2-0.04^2)$ $\boxed{720}$
Question Bank: w17
MSTE - Statistics and Probability / Probability / MSTE May 2019
Players $A$ and $B$ match pennies $N$ times. They keep a tally of their gains and losses. After the first loss, what is the chance that at no time during the game will they be even?
$\binom{N}{n}/2n$
$\binom{N}{n}/2^{n}$
$\binom{N}{n}/2$
$\binom{N}{n}/2^{N}$
The probability that the running tally is never even (tied) during the game is $\boxed{\dfrac{\binom{N}{n}}{2^{N}}}$
Question Bank: w232
MSTE - Statistics and Probability / Probability / MSTE April 2025
A spider eats three flies a day. Until he fills his quota, he has an even chance of catching any fly that attempts to pass. A fly is about to make the attempt. What are the chances of survival, given that five flies have already made the attempt today?
1/2
1/4
3/4
2/3
First find the probability that the quota of 3 flies is already complete after 5 attempts. Chance the first three flies were all captured $= (1/2)^{3} = 1/8$ Chance the 4th fly completed the quota $= \binom{3}{2}(1/2)^{2}(1/2)^{1}\times(1/2) = 3/16$ Chance the 5th fly completed the quota $= \binom{4}{2}(1/2)^{2}(1/2)^{2}\times(1/2) = 3/16$ $P(\text{quota complete}) = \dfrac{1}{8} + \dfrac{3}{16} + \dfrac{3}{16} = \dfrac{1}{2}$
For the 6th fly: Chance of being attacked $= 1 - \dfrac{1}{2} = \dfrac{1}{2}$ (the spider only attacks if the quota is not yet filled) Chance of being captured $= \left(\dfrac{1}{2}\right)\left(\dfrac{1}{2}\right) = \dfrac{1}{4}$ $P(\text{survival}) = 1 - \dfrac{1}{4}$ $\boxed{P = \tfrac{3}{4}}$
Question Bank: w243
MSTE - Statistics and Probability / Probability / MSTE April 2025
If a pair of dice is tossed 6 times, what is the probability of obtaining a total of 7 or 11 twice, a matching pair once, and any other combination 3 times?
0.0019
0.0129
0.1127
0.2156
Total number of outcomes per toss $= 6^{2} = 36$.
Outcome is 7 or 11: {1,6} {6,1} {2,5} {5,2} {3,4} {4,3} {5,6} {6,5} $P_1 = 8/36$
Outcome is a matching pair: {1,1} {2,2} {3,3} {4,4} {5,5} {6,6} $P_2 = 6/36$
Outcome is any other combination: $N = 36 - 8 - 6 = 22$, so $P_3 = 22/36$
Apply the multinomial probability for 6 tosses split 2–1–3: $P = P_1^{2}\cdot P_2^{1}\cdot P_3^{3}\cdot\dfrac{6!}{2!\,1!\,3!}$ $\boxed{P = 0.1127}$
Question Bank: w254
MSTE - Statistics and Probability / Probability / MSTE April 2025
A coin is so unbalanced that it may come both heads in 2 tosses as it may come tails in a single toss. What is the probability of getting a head in a single toss?
0.5
0.6
0.618
0.518
Let $h$ = probability of getting a head in 1 toss, so $1 - h$ = probability of getting a tail in 1 toss.
The problem states the two events are equally likely: $P(2\ \text{heads in 2 tosses}) = P(\text{tail in 1 toss})$ $h\cdot h = 1 - h$ $h^{2} + h - 1 = 0$ $h = 0.618$ and $-1.618$
Discarding the negative root (a probability cannot be negative): $\boxed{h = 0.618}$
Question Bank: w257
MSTE - Statistics and Probability / Probability / MSTE April 2025
If 2 marbles are removed at random from a bag containing black and white marbles, the chance that they are both white is 1/3. If 3 are removed at random, the chance that they all are white is 1/6. How many black balls are there?
6
7
5
4
Let $w$ = number of whites and $b$ = number of blacks, so the total is $w + b$.
Relate the two given probabilities: $W_1W_2W_3 = \dfrac{1}{6}$, and $W_1W_2 = \dfrac{1}{3}$ $\left(\dfrac{1}{3}\right)W_3 = \dfrac{1}{6}\ \Rightarrow\ W_3 = \dfrac{1}{2}$
$W_3$ is the chance the third draw is white after two whites are gone: $\dfrac{w-2}{w+b-2} = \dfrac{1}{2}\ \Rightarrow\ b = w - 2$, or $w = b + 2$
Substitute into the two-marble condition: $\dfrac{w}{w+b}\cdot\dfrac{w-1}{w+b-1} = \dfrac{1}{3}$ With $w = b+2$, solving gives $w = 6$ and $\boxed{b = 4}$
Question Bank: w265
MSTE - Statistics and Probability / Probability / MSTE April 2025
The local weather forecaster says "no rain" and his record is 2/3 accuracy of prediction. But the Federal Meteorological Service predicts "rain" and their record is 3/4. With no other data available, what is the chance of rain?
3/5
1/4
1/6
5/12
Rain happens when the local forecaster is wrong and the Federal service is right; no rain happens when the local forecaster is right and the Federal service is wrong.
Exam trap: stopping at $1/4$ gives choice B, which is the joint probability, not the conditional chance of rain.
Question Bank: w271
MSTE - Statistics and Probability / Probability / MSTE April 2025
In November 1994, Intel announced that a "subtle flaw" in its Pentium chip would affect 1 in 9 billion division problems. Suppose a computer performs 20 million divisions in the course of a particular program. What is the probability of no error? Of at least one error?
0.00000000111, 0.00222
0.99778, 0.00222
0.99999999989, 0.00000000111
0.00000000111, 0.99999999989
For a single operation: $q = \dfrac{1}{9\times 10^{9}}$ (probability of an error) $p = 1 - q$ (probability of no error)
For 20 million independent divisions: $P_{\text{no error}} = p^{20\times 10^{6}} = 0.99778$ $Q_{\text{at least 1 error}} = 1 - P_{\text{no error}} = 0.00222$
$\boxed{0.99778,\ 0.00222}$
Question Bank: w289
MSTE - Statistics and Probability / Probability / MSTE April 2025
If a chord is selected at random on a fixed circle what is the probability that its length exceeds the radius of the circle? Assume that the midpoint of the chord is evenly distributed over the circle.
0.5
0.667
0.866
0.75
A chord equals the radius when its midpoint lies on a concentric circle of radius $r = R\cos 30^\circ$. The chord is longer than the radius when its midpoint falls inside that inner circle.
With the midpoint uniformly distributed over the area of the circle, the probability is the ratio of areas: $P = \dfrac{\pi r^{2}}{\pi R^{2}} = \dfrac{(R\cos 30^\circ)^{2}}{R^{2}} = \cos^{2}30^\circ$ $\boxed{P = 0.75}$
This is Bertrand's paradox — the answer depends on how "at random" is defined: • midpoint uniform over the area: $P = \cos^{2}30^\circ = 0.75$ • distance from center uniform along a radius: $P = \cos 30^\circ = 0.866$ • endpoints uniform over the circumference: $P = 4/6 = 0.667$
Question Bank: w300
MSTE - Statistics and Probability / Probability / MSTE April 2025
A magician holds one six-sided die in his left hand and two in his right. What is the probability the number on the dice in his left hand is greater than the sum of the dice in his right?
5/54
2/17
7/108
1/9
The two right-hand dice have 36 equally likely outcomes. Count how many give a sum less than each possible left-hand value.
Left = 1, 2: impossible (the minimum right-hand sum is 2) Left = 3: sum of 2 → 1 way Left = 4: sums of 2 or 3 → 3 ways Left = 5: sums of 2 to 4 → 6 ways Left = 6: sums of 2 to 5 → 10 ways
MSTE - Statistics and Probability / Probability / MSTE April 2025
Situation 1. From where he stands, one step toward the cliff would send the drunken man over the edge. He takes random steps, either toward or away from the cliff. At any step his probability of taking a step away is 2/3, of a step toward the cliff 1/3.
What is his chance of escaping the cliff in exactly 3 steps?
16/27
8/243
2/27
1/2
He is one step from the edge. "Escaping in exactly 3 steps" means he is 3 steps clear after 3 moves, which happens either by three straight steps away, or by one step toward and two away in the orders that never take him over the edge.
MSTE - Statistics and Probability / Probability / MSTE April 2025
Situation 1. From where he stands, one step toward the cliff would send the drunken man over the edge. He takes random steps, either toward or away from the cliff. At any step his probability of taking a step away is 2/3, of a step toward the cliff 1/3.
What is his chance of falling in exactly 5 steps?
16/27
8/243
2/27
1/2
To fall on exactly the 5th step he must survive the first four, which requires two steps away and two steps toward the cliff arranged so he never goes over early, then a final step toward the cliff.
MSTE - Statistics and Probability / Probability / MSTE April 2025
Situation 1. From where he stands, one step toward the cliff would send the drunken man over the edge. He takes random steps, either toward or away from the cliff. At any step his probability of taking a step away is 2/3, of a step toward the cliff 1/3.
What is his chance of escaping the cliff?
16/27
8/243
2/27
1/2
Escaping means escaping in 1 step, or 2 steps, or 3 steps, … to infinity. For a biased random walk with an absorbing barrier, the escape probability is $P = 1 - \left(\dfrac{q}{p}\right)^{n}$ where $q$ = probability of a step toward the cliff $= 1/3$ $p$ = probability of a step away $= 2/3$ $n$ = initial position (number of steps from the edge) $= 1$
MSTE - Statistics and Probability / Probability / MSTE April 2025
How is Jane prepared for her test? Well, from past experience she thinks there is a probability of 0.1 that she will get an A, a probability of 0.4 that she will get a B, and a probability of 0.3 that she will get a C. What is the probability that she will get at least a C?
0.8
0.7
0.6
0.9
The event "at least a C" includes getting an A, a B, or a C. These are mutually exclusive, so the probabilities simply add.
$P(A) = 0.1$ $P(B) = 0.4$ $P(C) = 0.3$
$P(A \cup B \cup C) = P(A) + P(B) + P(C)$ $P(A \cup B \cup C) = 0.1 + 0.4 + 0.3$ $\boxed{P(\text{at least a C}) = 0.8}$
Question Bank: w315
MSTE - Statistics and Probability / Probability / MSTE April 2025
Find the reliability of the system shown, where A (0.95) and B (0.99) are in series, followed by C, D, and E (0.70 each) in parallel, then F and G (0.75 each) in parallel, and finally H (0.90) in series.
0.272
0.277
0.722
0.772
Reliability is the probability a unit functions within specified limits for at least a specified period under specified environment conditions.
Series system — the entire system fails if any component fails. Product Law of Reliabilities: $R_s = \prod_{i=1}^{n} R_i$
Parallel system — the system fails only when all components fail. Product Law of Unreliabilities: $R_p = 1 - \prod_{i=1}^{n}(1 - R_i)$
