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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 marked chance

The figure places event A on a probability scale. Write its probability as a percent.

Event A placed on a probability scale from $0$ to $1$ .

Text description of this figure

A horizontal number line runs from 0 at the left end to 1 at the right end. Six evenly spaced tick marks divide it into five equal intervals, and only the two ends are labeled, 0 and 1; the four interior ticks carry no numbers. A filled dot sits on the fourth tick to the right of 0 and is labeled A above the line.
1. Hint 1
  The distance from $0$ to $1$ represents the whole chance, or $100\%$ .
2. Hint 2
  Use the equal intervals to find the fraction of the scale reached by A, then convert it to a percent.
Answer

$80\%$ .

Full solution

The scale divides the whole into five equal intervals, and A is four intervals from $0$ .
Its probability is $\frac45$ .
$\frac45=\frac{80}{100}.$

That is $80\%$ , which lies between an even chance and certainty.

Answer

$80\%$ .

Key idea

A probability scale locates a share of one whole, which can also be written as a percent.
Problem 1: Mark complete
Problem 2 Marked tickets

Exactly $40$ tickets are equally likely to be selected. Some are marked, and the probability of selecting a marked ticket is $0.35$ . How many tickets are marked?
1. Hint 1
  The given probability is the marked share of all the tickets.
2. Hint 2
  Multiply the total number of tickets by that share.
Answer

$14$ tickets.

Full solution

Equal likelihood makes the probability the number marked divided by $40$ .
Therefore the marked count is
$0.35\times40=14.$

Check: $\frac{14}{40}=0.35$ , as required.

Answer

$14$ tickets.

Key idea

With equally likely outcomes, multiplying a probability by the total count recovers the favorable count.
Problem 2: Mark complete
Problem 3 A trial record

An event occurred $18$ times in a completed experiment. Its experimental probability was $0.45$ . How many trials were recorded?
1. Hint 1
  Experimental probability compares the occurrence count with the trial total.
2. Hint 2
  If $0.45$ of the trials account for $18$ occurrences, divide to recover the whole trial count.
Answer

$40$ trials.

Full solution

The occurrence count is $0.45$ times the number of trials.
Dividing gives
$18\div0.45=40.$

Check: $\frac{18}{40}=0.45$ , so $40$ trials fit the record.

Answer

$40$ trials.

Key idea

The occurrence count divided by the experimental probability recovers the number of trials when that probability is nonzero.
Problem 3: Mark complete
Problem 4 Selecting a cell

The grid shows a board whose cells are chosen by selecting a row and a column. Each row is equally likely, each column is equally likely, and the two selections are independent. What is the probability that the selected row label is greater than the selected column label? Give a fraction in simplest form.

The board, with its row labels and column labels.

Text description of this figure

A grid of three equal rows and four equal columns, making twelve cells, every one of them blank. Down the left side, under the heading Row label, the rows are labeled 2, 4 and 6 from top to bottom. Across the top, under the heading Column label, the columns are labeled 1, 3, 5 and 7 from left to right.
1. Hint 1
  Each possible row and column pair identifies one cell.
2. Hint 2
  In each row, count the column labels smaller than that row label.
3. Hint 3
  Add the favorable cell counts and compare them with all cells.
Answer

$\frac12$ .

Full solution

The independent selections give equally likely cells.
There are three rows and four columns, so
$3\times4=12.$

Row $2$ has one smaller column label, $1$ .
Row $4$ has two, $1$ and $3$ .
Row $6$ has three, $1$ , $3$ , and $5$ .
Thus the favorable count is
$1+2+3=6.$

The probability is
$\frac{6}{12}=\frac12.$

No row label equals a column label, so the other six cells all have a smaller row label, and the two groups fill the board.

Answer

$\frac12$ .

Key idea

Favorable cells over all cells of an outcome grid give the probability when each stage is equally likely and the stages are independent.
Problem 4: Mark complete
Problem 5 The image collection

A collection contains $12$ images. Each is equally likely to be selected, and the probability of selecting a portrait is $\frac13$ . Three new portraits are added, with no images removed. Every image in the enlarged collection is equally likely to be selected. What is the new probability of a portrait? Give a fraction in simplest form.
1. Hint 1
  First find how many portraits were in the original collection.
2. Hint 2
  The added portraits change both the favorable count and the total count.
3. Hint 3
  Use the new counts in the probability fraction.
Answer

$\frac7{15}$ .

Full solution

The original portrait count is
$\frac13\times12=4.$

Adding three portraits gives $4+3=7$ portraits among $12+3=15$ images.
Thus the new probability is
$P(\text{portrait})=\frac7{15}.$

The original eight images that were not portraits are still present, and $7+8=15$ checks the new counts.

Answer

$\frac7{15}$ .

Key idea

Adding favorable outcomes changes both parts of a probability fraction.
Problem 5: Mark complete
Problem 6 Two trial records

A device has theoretical probability $\frac14$ of flashing green on each trial. All trials are independent. One record shows $7$ green flashes in $25$ trials, and a separate record shows $11$ in $35$ trials. Find the experimental probability from the combined records as a decimal, and state whether it is above, below, or equal to the theoretical probability.
1. Hint 1
  The two records together form one longer run of the same device.
2. Hint 2
  Add the green flashes from both records, and add the trials from both records.
3. Hint 3
  Write the theoretical fraction as a decimal for the comparison.
Answer

$0.3$ ; above the theoretical probability.

Full solution

The records contain $7+11=18$ green flashes and $25+35=60$ trials.
$\frac{18}{60}=0.3.$

The theoretical value is
$\frac14=0.25.$

Since $0.3>0.25$ , the combined experimental probability is above the theoretical probability.
A trial record can differ from theory even when the device has the stated probability.

Answer

$0.3$ ; above the theoretical probability.

Key idea

When records come from the same experiment, pool their counts before comparing the relative frequency with the theoretical probability.
Problem 6: Mark complete
Problem 7 A robot journey

A robot independently chooses a turn, left or right, and a distance of $1$ , $2$ , $3$ , or $4$ meters. Each turn is equally likely, and each distance is equally likely. It runs out of battery exactly for these choices: left with $4$ meters, right with $3$ meters, and right with $4$ meters. For every other pair of choices it completes the journey. What is the probability it completes the journey? Give a fraction in simplest form.
1. Hint 1
  Completing the journey and running out of battery cover all possibilities without overlap.
2. Hint 2
  Count the equally likely turn and distance pairs, then find the probability that one of the listed failures occurs.
3. Hint 3
  Subtract the failure probability from one.
Answer

$\frac58$ .

Full solution

There are two turn choices and four distance choices.
Equal likelihood at each stage and independence make all pairs equally likely.
$2\times4=8.$

The three listed failures are different outcomes, so their probability is $\frac38$ .
The completion probability is
$1-\frac38=\frac58.$

Check directly: completion allows three left distances and two right distances, giving five of the eight outcomes.

Answer

$\frac58$ .

Key idea

Subtracting the probability of the outcomes that fail gives the probability of completion.
Problem 7: Mark complete
Problem 8 The running record

A fair chooser produces A, B, C, or D with equal probability on independent trials. A occurs twice in the first $8$ trials, then occurs again on trial $9$ . Omar says the experimental probability of A is now closer to the theoretical probability. Is his claim correct? Justify your answer.
1. Hint 1
  Find the theoretical probability and the two experimental probabilities.
2. Hint 2
  Compare the old record with theory before deciding whether the new record is closer.
Answer

No; the value changes from $\frac14$ to $\frac13$ , farther from the theoretical $\frac14$ .

Full solution

One of four equally likely outputs is A, so the theoretical probability is $\frac14$ .
Before trial nine, the experimental probability was
$\frac28=\frac14.$

After another A, there are three occurrences in nine trials.
$\frac39=\frac13.$

The old value matched theory exactly, while the new value differs from it.
Omar's claim is false: one more independent trial has moved the experimental probability farther from theory.

Answer

No; the value changes from $\frac14$ to $\frac13$ , farther from the theoretical $\frac14$ .

Key idea

One additional independent trial may move an experimental probability farther from its theoretical value.
Problem 8: Mark complete
Problem 9 Two trial plans

A device gives a red signal with theoretical probability $0.4$ each time it runs. Separate runs of the device do not affect one another. Plan A runs the device $1{,}000$ times and records each run's own result. Plan B runs it once and writes that result into all $1{,}000$ record entries. Which plan provides a reason to expect its recorded red fraction to be close to $0.4$ ? Explain, and state all possible red fractions for Plan B.
1. Hint 1
  Many entries in a record need not represent many independent trials.
2. Hint 2
  Count how many actual runs of the device each record contains.
3. Hint 3
  In Plan B, decide how many red entries follow a red result and how many follow a result that is not red.
Answer

Plan A; Plan B can record red fractions of $0$ or $1$ .

Full solution

Plan A supplies many independent trials.
The law of large numbers gives a reason to expect its experimental red fraction to lie close to $0.4$ , although it does not guarantee a close result in this particular run.

Plan B supplies just one actual trial.
If that result is red, every copied entry is red.
The recorded fraction is
$\frac{1{,}000}{1{,}000}=1.$
If it is not red, no copied entry is red.
The recorded fraction is
$\frac{0}{1{,}000}=0.$

These are the two possible fractions for Plan B.
Copying the result adds no independent trials, so the larger record does not support the same expectation of closeness.

Answer

Plan A; Plan B can record red fractions of $0$ or $1$ .

Key idea

A long record supports a close experimental estimate through independent trials, not through copies of one outcome.
Problem 9: Mark complete
Problem 10 Choosing a file

A program first chooses folder X or folder Y, with equal probability. Folder X contains one file, and folder Y contains three files. It then selects one file from the chosen folder, with each file in that folder equally likely. Pat says the file in X is more likely to be selected than any one file in Y. Is Pat correct? Give the probability of selecting the file in X and explain.
1. Hint 1
  Equal chances for the folders do not automatically give equal chances to every file across both folders.
2. Hint 2
  Ask which first choice guarantees that the file in X will be selected.
3. Hint 3
  A file in Y needs folder Y to be chosen first, and then needs to be the one file picked from three.
Answer

Yes; the file in X has probability $\frac12$ .

Full solution

The file in X is selected exactly when the first choice is folder X.
There are two equally likely folder choices, so
$P(\text{file in X})=\frac12.$

A particular file in Y can be selected only when the first choice is folder Y, which also has probability $\frac12$ .
Even then, it is selected only if it is the one file picked from the three in Y, and each of the other two files in Y can be picked instead.
So its probability is less than $\frac12$ .

Therefore the file in X is more likely to be selected than any one file in Y, and Pat is correct.

The four files do not have equal chances, so counting one file out of four would give the wrong value $\frac14$ .

Answer

Yes; the file in X has probability $\frac12$ .

Key idea

Counting favorable outcomes over all outcomes requires equal likelihood for the outcomes being counted.
Problem 10: Mark complete

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