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Introduction to Probability: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 The overlapping groups

One of the 50 outcomes represented in the figure is selected with equal probability. Find $P(A\cap B)$ .

Outcome counts in the sample space S.

Text description of this figure

A rectangle labeled S contains two overlapping circles, A on the left drawn solid and B on the right drawn dashed. Each region holds a count of outcomes: 15 in the part of A that is outside B, 5 in the lens where A and B overlap, 10 in the part of B that is outside A, and 20 in the corner of the rectangle outside both circles. No probability is written.
1. Hint 1
  The intersection consists of outcomes belonging to both events.
2. Hint 2
  Read the overlap count and divide by the total count in the sample space.
Answer

$\frac{1}{10}$ .

Full solution

The overlap contains $5$ outcomes out of $50$ equally likely outcomes.
Thus
$\begin{aligned}P(A\cap B)&=\frac5{50} \\ P(A\cap B)&=\frac1{10}.\end{aligned}$
The four regions sum to $15+5+10+20=50$ , confirming the total.

Answer

$\frac{1}{10}$ .

Key idea

An intersection contains the outcomes shared by both events.
Problem 1: Mark complete
Problem 2 Two equally likely events

Events A and B have equal probability. Their intersection has one-fourth the probability of A, and their union has probability $0.7$ . Find $P(A)$ .
1. Hint 1
  Adding the two event probabilities counts their overlap twice.
2. Hint 2
  Write both event probabilities using one variable, and express the overlap as a fraction of that variable.
Answer

$P(A)=0.4$ .

Full solution

Let $p$ be the common value of $P(A)$ and $P(B)$ .
The intersection then has probability $p/4$ .
Subtracting the double-counted overlap gives
$2p-\frac p4=0.7.$
Thus $(7/4)p=0.7$ , so $p=0.4$ .
The overlap is $0.1$ , and the union checks as
$0.4+0.4-0.1=0.7.$

Answer

$P(A)=0.4$ .

Key idea

Relations among event probabilities can be combined with the overlap correction to determine an unknown probability.
Problem 2: Mark complete
Problem 3 An event share

You know $P(B)>0$ , $P(A\mid B)=0.30$ , and $P(A\cap B)=0.15$ . Find $P(B)$ .
1. Hint 1
  The conditional probability measures the fraction of B that also belongs to A.
2. Hint 2
  Use $P(A\cap B)=P(B)P(A\mid B)$ and solve for the missing probability.
Answer

$P(B)=0.50$ .

Full solution

The multiplication rule gives
$0.15=0.30P(B).$
Divide by the nonzero number $0.30$ :
$P(B)=0.50.$
Then $0.15/0.50=0.30$ , checking the conditional probability.

Answer

$P(B)=0.50$ .

Key idea

A joint probability and a nonzero conditional probability determine the conditioning event's probability.
Problem 3: Mark complete
Problem 4 Two backup devices

Two backup devices work independently. The first works with probability $0.8$ , and the second with probability $0.6$ . A task succeeds if at least one device works. Find its success probability.
1. Hint 1
  The task fails exactly when both devices fail.
2. Hint 2
  Independence also lets you multiply the two failure probabilities.
Answer

$0.92$ .

Full solution

The failure probabilities are $0.2$ and $0.4$ .
Independence gives
$\begin{aligned}P(\text{both fail})&=0.2(0.4) \\ P(\text{both fail})&=0.08.\end{aligned}$
Taking the complement,
$\begin{aligned}P(\text{success})&=1-0.08 \\ P(\text{success})&=0.92.\end{aligned}$
The direct disjoint cases, first works or first fails and second works, give $0.8+0.2(0.6)=0.92$ as a check.

Answer

$0.92$ .

Key idea

Independent devices can have different probabilities, and total failure is the complement of at least one success.
Problem 4: Mark complete
Problem 5 Two chosen letters

Two different letters are chosen in order from A, E, B, and C, with every ordered pair equally likely. Let J mean the first letter is A, and let V mean the second is a vowel, A or E. Find $P(V\mid J)$ and $P(J\cap V)$ , and determine whether J and V are independent.
1. Hint 1
  Once the first letter is known, remove it from the available second letters.
2. Hint 2
  Compare the conditional chance of V after J with its original chance, or compare the joint probability with the product.
Answer

$P(V\mid J)=\frac13$ ; $P(J\cap V)=\frac1{12}$ ; J and V are dependent.

Full solution

Given J, the second letter is equally likely to be E, B, or C, with one vowel, so
$P(V\mid J)=\frac13.$
The first letter is A with probability $\frac14$ , so
$\begin{aligned}P(J\cap V)&=\frac14\cdot\frac13 \\ P(J\cap V)&=\frac1{12}.\end{aligned}$
This matches the single qualifying pair AE among the $12$ ordered pairs.

Across all ordered pairs, each letter is equally likely in the second position, giving
$P(V)=\frac12.$
The conditional probability $\frac13$ differs from it, and $P(J)P(V)=\frac18$ differs from $\frac1{12}$ , so the events are dependent.

Answer

$P(V\mid J)=\frac13$ ; $P(J\cap V)=\frac1{12}$ ; J and V are dependent.

Key idea

Removing a specific object can change the probability of a broader category on the next choice.
Problem 5: Mark complete
Problem 6 Downloaded audio

For a randomly chosen file, $40\%$ are audio and $25\%$ are downloaded. Every downloaded file is audio. Find the probability a downloaded file is audio and the probability an audio file is downloaded.
1. Hint 1
  The downloaded event is a subset of the audio event.
2. Hint 2
  The intersection is the downloaded event; divide it by the probability of each conditioning event in turn.
Answer

Audio given downloaded: $1$ ; downloaded given audio: $\frac58$ .

Full solution

Let A mean audio and D mean downloaded.
Every downloaded file is audio, so $P(A\cap D)=0.25$ , and both conditioning probabilities are positive.
Thus
$\begin{aligned}P(A\mid D)&=\frac{0.25}{0.25} \\ P(A\mid D)&=1.\end{aligned}$
Reversing the condition gives
$\begin{aligned}P(D\mid A)&=\frac{0.25}{0.40} \\ P(D\mid A)&=\frac58.\end{aligned}$
The two conditions refer to different sample spaces.

Answer

Audio given downloaded: $1$ ; downloaded given audio: $\frac58$ .

Key idea

Reversing a conditional probability changes which event supplies the denominator.
Problem 6: Mark complete
Problem 7 A survey reconstruction

Of 80 visitors, 40 tried activity A, 30 tried activity B, and 20 tried neither. A visitor is selected uniformly at random. Find the probability that the visitor tried B given that the visitor tried A.
1. Hint 1
  First recover the number who tried at least one activity.
2. Hint 2
  The two activity counts exceed that total by the number who tried both.
Answer

$\frac14$ .

Full solution

The union contains $80-20=60$ visitors.
Adding the activity counts counts the overlap twice, so the overlap is
$40+30-60=10.$
Among the $40$ who tried A, $10$ also tried B.
Therefore
$\begin{aligned}P(B\mid A)&=\frac{10}{40} \\ P(B\mid A)&=\frac14.\end{aligned}$
The regions are $30$ A only, $10$ both, $20$ B only, and $20$ neither, totaling $80$ .

Answer

$\frac14$ .

Key idea

Recover the overlap before restricting a probability to one event.
Problem 7: Mark complete
Problem 8 An event and its opposite

An event A has probability $0.2$ . A student says A and its complement are independent because they are different events. Is this correct? Justify your answer numerically.
1. Hint 1
  An event and its complement cannot occur together.
2. Hint 2
  Compare their joint probability with the product of their separate probabilities.
Answer

No; A and its complement are dependent: $P(A\cap A^c)=0$ , whereas $P(A)P(A^c)=0.16$ .

Full solution

The complement has probability $0.8$ , and the two events have no common outcomes.
Thus $P(A\cap A^c)=0$ , while
$\begin{aligned}P(A)P(A^c)&=0.2(0.8) \\ P(A)P(A^c)&=0.16.\end{aligned}$
The two values differ, so independence fails.
In fact, knowing that one happened rules out the other.

Answer

No; A and its complement are dependent: $P(A\cap A^c)=0$ , whereas $P(A)P(A^c)=0.16$ .

Key idea

An event with probability strictly between zero and one is dependent on its complement.
Problem 8: Mark complete
Problem 9 Enough information?

Events A and B satisfy $P(A)=4/11$ and $P(B)=6/11$ . Do these probabilities determine whether the events are independent? Give two possible intersection probabilities, one making the events independent and one making them dependent.
1. Hint 1
  Independence depends on the intersection as well as on the individual probabilities.
2. Hint 2
  Find the overlap independence requires, then look for a different overlap compatible with both events.
Answer

No. An intersection probability of $\frac{24}{121}$ gives independence; an intersection probability of $0$ gives dependence.

Full solution

Independence requires $P(A\cap B)=\frac4{11}\cdot\frac6{11}=\frac{24}{121}$ .
This is possible: among $121$ equally likely outcomes, put $24$ in both events, $20$ in A only, $42$ in B only and $35$ in neither, so that A has $44$ and B has $66$ outcomes.

Dependence is also possible: among $11$ equally likely outcomes, put $4$ in A only, $6$ in B only and $1$ in neither.
The intersection then has probability $0$ , which differs from the product, so the events are dependent.

Answer

No. An intersection probability of $\frac{24}{121}$ gives independence; an intersection probability of $0$ gives dependence.

Key idea

Unless an event has probability 0 or 1, the two events' probabilities alone do not decide whether they are independent.
Problem 9: Mark complete
Problem 10 A two-stage selection

Three boxes contain 1 red object, 2 blue objects, and 3 green objects, respectively. A box is selected uniformly, then an object is selected uniformly from that box. A student says the red object has probability $1/6$ because there are six objects. Is this correct? Find its actual selection probability.
1. Hint 1
  Equal probabilities for boxes do not imply equal probabilities for all individual objects.
2. Hint 2
  For the red object to be selected, first its box must be selected; that box offers one possible object.
Answer

No; the red object has probability $\frac13$ .

Full solution

The red box is selected with probability $1/3$ .
Given that box, the red object is certain, so
$\begin{aligned}P(\text{red object})&=\frac13\cdot1 \\ P(\text{red object})&=\frac13.\end{aligned}$
Each blue object instead has probability $1/6$ , and each green object $1/9$ .
The six objects are not equally likely, so dividing by six was invalid.

Answer

No; the red object has probability $\frac13$ .

Key idea

A uniform choice at each stage does not necessarily make all final objects equally likely.
Problem 10: Mark complete

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