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Data, Counting, and Probability: Star problems

Ten optional challenges to stretch your reasoning. Work on paper, use hints when you need them, and check the answer or full solution when you are ready. You can skip these problems and continue the course.

1 of 3 stars: Stretch
2 of 3 stars: Challenge
3 of 3 stars: Deep challenge

Stars indicate difficulty within this set.

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Problem 1 of 10

Problem 1 The two missing extremes

Difficulty: 1 of 3 stars, Stretch

Ten whole-number scores have a mean of 18. After one smallest score and one largest score are removed, the mean of the remaining eight is 19. The range of the original ten scores is 14.

Find the removed scores. Then give a complete list of ten scores showing that all the conditions can hold at once.
1. Hint 1
  Convert each mean into a total before comparing the two groups.
2. Hint 2
  You know both the sum and the difference of the removed scores. Split their sum into two equal parts first.
Answer

The removed scores are 7 and 21. One valid list is 7, eight copies of 19, and 21.

Full solution

The total of the original ten scores is
$10\times18=180.$
The eight remaining scores total
$8\times19=152.$
Thus the two removed scores have sum
$180-152=28.$

Because they were a smallest and a largest score, their difference is the original range, 14.
Two equal numbers with sum 28 would both be 14.
To make their difference 14 while keeping the sum fixed, move 7 from one to the other.
The resulting numbers are $14-7=7$ and $14+7=21$ .

Finding these endpoints is not quite enough: the other eight scores must fit between them and meet the stated mean.
Choose all eight to be 19.
The full list then has total $7+8\times19+21=180$ , range $21-7=14$ , and original mean 18.
Removing the smallest and largest leaves eight 19s, whose mean is 19.
This construction verifies that the numerical deductions describe an actual possible data set.

Answer

The removed scores are 7 and 21. One valid list is 7, eight copies of 19, and 21.

Key idea

A mean fixes a total; a construction checks whether derived statistics are jointly possible.
Problem 1: Mark complete
Problem 2 Mean, median, and range together

Difficulty: 1 of 3 stars, Stretch

A data set consists of five distinct positive whole numbers. Its mean is 8, its median is 7, and its range is 10.

Find every possible data set. Write each in increasing order and justify that none is missing.
1. Hint 1
  Write the ordered numbers as $a,b,7,d,a+10$ . Their total is 40.
2. Hint 2
  The number $b$ is at most 6, while $d$ is at most $a+9$ . Lower bounds on $b$ and $d$ are useful too.
Answer

The four data sets are $(3,5,7,12,13)$ , $(3,6,7,11,13)$ , $(4,5,7,10,14)$ , and $(4,6,7,9,14)$ .

Full solution

Let the ordered numbers be
$a<b<7<d<a+10.$
The final term follows from the range of 10, and the middle term follows from the median.
The total is $5\times8=40$ , so $b+d=23-2a$ .

Since the numbers are distinct whole numbers, $b\leq6$ and $d\leq a+9$ .
Thus
$b+d\leq a+15.$
If $a$ were 1 or 2, the required value $23-2a$ would exceed that upper bound.
Hence $a\geq3$ .

Also $b\geq a+1$ and $d\geq8$ , so
$b+d\geq a+9.$
For $a\geq5$ , this is greater than the required $23-2a$ .
Thus $a$ is either 3 or 4.

For $a=3$ , the sum $b+d$ is 17.
The possible $b$ values are 4, 5, and 6; the first gives $d=13$ , equal to the largest number and therefore forbidden.
The others give the first two listed data sets.
For $a=4$ , the sum is 15, and $b=5$ or 6 gives the other two.
Each listed set is distinct, totals 40, has median 7, and has range 10.

Answer

The four data sets are $(3,5,7,12,13)$ , $(3,6,7,11,13)$ , $(4,5,7,10,14)$ , and $(4,6,7,9,14)$ .

Key idea

Combining summary statistics can sharply restrict the data, but completeness needs bounds.
Problem 2: Mark complete
Problem 3 The missing graph scale

Difficulty: 1 of 3 stars, Stretch

A bar chart shows counts for five days, in order. Measured upward from the chart's displayed baseline, the bar heights are 1, 4, 2, 5, and 3 equal grid intervals. The vertical scale uses the same count increase for every grid interval, but all numerical scale labels are missing. The displayed baseline is not necessarily zero.

The five counts total 95, and their range is 12.

(a) Recover the five counts in day order and the value represented by the displayed baseline.

(b) A student says the tallest bar represents five times the count of the shortest bar. Is the claim correct? Explain.

Text description of this figure

A bar chart with five bars labeled Day 1 to Day 5 from left to right. Every bar starts on a thick horizontal line labeled Displayed baseline, value unknown. Five equally spaced horizontal gridlines lie above the baseline, and a heading above the chart reads Equal vertical grid intervals. The vertical axis carries no numbers. Counted in grid intervals above the baseline, the bars reach heights of 1, 4, 2, 5 and 3.
1. Hint 1
  The tallest and shortest bars differ by four grid intervals.
2. Hint 2
  The average plotted height is three grid intervals above the baseline. The average count is $95/5$ .
Answer

Counts: 13, 22, 16, 25, 19. Displayed baseline: 10. The claim is false; the actual ratio is $25/13$ .

Full solution

The range of the plotted heights is $5-1=4$ grid intervals.
Since the actual range is 12, one interval represents an increase of $12/4=3$ in the count.

The five plotted heights total $1+4+2+5+3=15$ intervals, so their mean height is 3 intervals above the baseline.
The five actual counts have mean $95/5=19$ .
Three intervals represent 9 counts; therefore the baseline represents $19-9=10$ .

Add the appropriate multiples of 3 to 10.
The five counts are $10+3=13$ , $10+12=22$ , $10+6=16$ , $10+15=25$ , and $10+9=19$ .
They total 95, and $25-13=12$ , confirming both given statistics.

The visible heights have ratio 5 to 1, but those heights measure the amounts above 10, not the full counts.
The actual tallest-to-shortest ratio is $25/13$ , which is less than 2.
Multiplying a visible height does not multiply the full value when the baseline is nonzero.

Answer

Counts: 13, 22, 16, 25, 19. Displayed baseline: 10. The claim is false; the actual ratio is $25/13$ .

Key idea

Differences survive a shifted graph baseline, but ratios generally do not.
Problem 3: Mark complete
Problem 4 Two forbidden intersections

Difficulty: 2 of 3 stars, Challenge

A robot travels on a square grid from $(0,0)$ to $(5,3)$ . Each move is either one unit right or one unit up. It may not visit either $(2,1)$ or $(3,2)$ .

How many different allowed routes are there? Two routes are different if their sequences of moves differ. Explain how your count avoids both omissions and double counting.

Text description of this figure

A square grid 5 units wide and 3 units tall, numbered 0 to 5 along the bottom and 1 to 3 up the left side. A dot at the bottom left corner is labeled Start, and a dot at the top right corner, 5 across and 3 up, is labeled Finish. Two crosses mark the forbidden points: one at 2 across and 1 up, the other at 3 across and 2 up. A note beside the grid reads: Crosses mark forbidden points.
1. Hint 1
  First count every route, then count routes through each forbidden point.
2. Hint 2
  Routes through both forbidden points would be subtracted twice. Add those routes back once.
Answer

There are 14 allowed routes.

Full solution

Every unrestricted route has eight moves: five right and three up.
Choose the three positions occupied by up moves.
There are $8\times7\times6$ ordered selections of three different positions, but each set of positions is counted $3\times2\times1=6$ times.
Thus there are $8\times7\times6/6=56$ routes.

Call the forbidden points $A=(2,1)$ and $B=(3,2)$ .
There are 3 routes to $A$ , because its one up move can occupy any of three positions.
From $A$ to the finish, choose two up positions among five moves: $5\times4/2=10$ ways.
Therefore 30 routes visit $A$ .

There are $5\times4/2=10$ routes to $B$ and 3 routes from $B$ to the finish, so 30 routes visit $B$ .
Some routes occur in both counts.
Such a route must visit $A$ first, because right-and-up moves cannot return from $B$ to $A$ .
It has 3 choices to reach $A$ , 2 choices from $A$ to $B$ , and 3 choices afterward: $3\times2\times3=18$ routes.

Subtracting both forbidden counts removes those 18 routes twice.
Adding them back once gives
$56-30-30+18=14.$
Each forbidden route is now removed exactly once, and every allowed route remains.

Answer

There are 14 allowed routes.

Key idea

When subtracting overlapping bad cases, restore the overlap once.
Problem 4: Mark complete
Problem 5 Two different filters

Difficulty: 2 of 3 stars, Challenge

A fair spinner has four equal sectors labeled 1, 2, 3, and 4. It is spun twice independently; the first and second results are recorded in order.

(a) Keep a trial only if at least one of its two results is 3. Among the kept trials, what is the probability that the sum is at least 6?

(b) Start again, and instead keep a trial only if its first result is 3. Among these kept trials, what is the probability that the sum is at least 6? Explain why the two answers need not agree.
1. Hint 1
  List ordered pairs. The pairs $(3,4)$ and $(4,3)$ are different outcomes.
2. Hint 2
  In part (a), the outcome $(3,3)$ belongs to both the first-is-3 and second-is-3 lists, but must be counted only once.
Answer

Part (a): $3/7$ . Part (b): $1/2$ .

Full solution

Before any filtering, the 16 ordered pairs are equally likely, because each spin has four equally likely results and the spins are independent.
Conditioning on a filter means keeping only the pairs that pass it and measuring the favorable fraction within that smaller set.

For part (a), the kept pairs are $(3,1),(3,2),(3,3),(3,4),(1,3),(2,3),(4,3)$ .
There are seven, not eight: $(3,3)$ cannot be listed twice.
Exactly three have sum at least 6: $(3,3),(3,4),(4,3)$ .
The probability is therefore $3/7$ .

For part (b), only $(3,1),(3,2),(3,3),(3,4)$ remain.
Two of these four have sum at least 6, giving probability $2/4=1/2$ .

The statements "at least one result is 3" and "the first result is 3" preserve different sets of trials.
Both tell us that a 3 occurred, but they supply different information about its position.
The correct denominator comes from the exact filter, not merely from the fact that one result is known to be 3.

Answer

Part (a): $3/7$ . Part (b): $1/2$ .

Key idea

Probability depends on the precise information used to select the sample.
Problem 5: Mark complete
Problem 6 Codes that are also numbers

Difficulty: 2 of 3 stars, Challenge

How many four-digit positive whole numbers can be formed from the digits 0, 1, 2, 3, 4, and 5 if no digit repeats, the number is divisible by 5, and exactly two of its digits are odd?

A four-digit number cannot begin with 0. Give a complete count without listing every number.
1. Hint 1
  Separate numbers ending in 0 from numbers ending in 5.
2. Hint 2
  In the case ending in 5, count all arrangements of the other digits first, then remove those that start with 0.
Answer

There are 64 numbers.

Full solution

A number divisible by 5 ends in 0 or 5.
These two cases do not overlap, so their counts can be added.

If the last digit is 0, the first three positions contain two odd digits from 1, 3, 5 and one even digit from 2, 4.
Choose the even digit's position in 3 ways and its value in 2 ways.
Fill the other two positions with different odd digits in $3\times2$ ways.
This gives $3\times2\times3\times2=36$ numbers.
None starts with 0, since 0 is already at the end.

If the last digit is 5, the first three positions need one odd digit from 1, 3 and two even digits from 0, 2, 4.
There are 2 odd choices and 3 even pairs: $(0,2),(0,4),(2,4)$ .
The three selected digits can be arranged in $3\times2\times1=6$ orders, giving 36 arrangements before checking the leading digit.

The invalid arrangements start with 0.
Choose the other even digit in 2 ways and the odd digit in 2 ways, then arrange those two in the remaining positions in 2 ways: 8 invalid arrangements.
This case contributes $36-8=28$ .
The complete count is $36+28=64$ .

Answer

There are 64 numbers.

Key idea

A restriction that changes with the last digit is easier to handle by separate cases.
Problem 6: Mark complete
Problem 7 A large mean with a fixed median and mode

Difficulty: 2 of 3 stars, Challenge

Seven scores are whole numbers from 0 through 10, inclusive. Their median is 6, and their unique mode is 8: the score 8 occurs more often than any other individual score.

What is the largest possible mean? Give a data set attaining it and prove that no larger mean is possible.
1. Hint 1
  Put the scores in increasing order. There are only three positions above the median.
2. Hint 2
  The score 8 must appear either twice or three times. Optimize the total separately in those two cases.
Answer

The maximum mean is $46/7$ , attained by $5,5,6,6,8,8,8$ .

Full solution

In increasing order, the fourth score is 6.
Therefore every 8 must occupy one of the final three positions, so 8 appears at most three times.
It must appear at least twice: if it appeared once, another score would also appear at least once, preventing it from being the unique mode.

If 8 appears three times, the last three scores are $8,8,8$ .
Each other value can appear at most twice.
The first four scores are at most 6, with the fourth equal to 6.
Their largest possible total is obtained by using two 6s and two 5s: 22.
To justify this bound, at most two entries can contribute 6, and each remaining entry is at most 5.
The full total is therefore at most $22+24=46$ .

If 8 appears twice, every other value must appear at most once.
The first four scores are distinct and at most 6, so their total is at most
$3+4+5+6=18.$
The remaining scores are two 8s and at most a 10, giving full total at most
$18+8+8+10=44.$

The first case is better.
The list $5,5,6,6,8,8,8$ has median 6, unique mode 8, and total 46.
Thus the maximum mean is $46/7$ .

Answer

The maximum mean is $46/7$ , attained by $5,5,6,6,8,8,8$ .

Key idea

To optimize a statistic, convert it to a total and separate the structural cases.
Problem 7: Mark complete
Problem 8 A bag with balanced outcomes

Difficulty: 3 of 3 stars, Deep challenge

A bag contains only red and blue counters, with at least one of each color and at most 20 counters in total. There are at least as many red counters as blue counters. Two counters are drawn uniformly at random without replacement.

The probability that the two counters have the same color equals the probability that they have different colors. Find every possible pair of red and blue counts, and prove that your list is complete.
1. Hint 1
  If there are $r$ red and $b$ blue counters, count ordered pairs of distinct individual counters.
2. Hint 2
  Equal same-color and different-color counts lead to $(r-b)^2=r+b$ . The total must therefore be a perfect square.
Answer

The possibilities are 3 red and 1 blue; 6 red and 3 blue; or 10 red and 6 blue.

Full solution

Let the counts be $r$ and $b$ , with
$r\geq b\geq1.$
Every ordered pair of distinct individual counters is equally likely.
Same-color pairs number $r(r-1)+b(b-1)$ .
Different-color pairs number $rb+br=2rb$ .
Equality of the probabilities is equivalent to equality of these counts.

Thus
$r^2-r+b^2-b=2rb.$
Rearranging gives $r^2-2rb+b^2=r+b$ , or
$(r-b)^2=r+b.$
Let $d=r-b$ , a nonnegative whole number.
The total is $d^2$ , and it is between 2 and 20.
Therefore $d$ can only be 2, 3, or 4.
The values 0 and 1 give too few counters; values at least 5 give at least 25.

For $d=2$ , total 4 and difference 2 give $r=3,b=1$ .
For $d=3$ , total 9 and difference 3 give $r=6,b=3$ .
For $d=4$ , total 16 and difference 4 give $r=10,b=6$ .

Each pair contains both colors, meets the total limit, and satisfies the original ordered-pair equality.
Since every possible difference was examined, these are all solutions.
Drawing without replacement matters: the terms are $r(r-1)$ and $b(b-1)$ , not $r^2$ and $b^2$ .

Answer

The possibilities are 3 red and 1 blue; 6 red and 3 blue; or 10 red and 6 blue.

Key idea

Counting equally likely outcomes can turn a probability condition into an integer classification.
Problem 8: Mark complete
Problem 9 Even rows and even columns

Difficulty: 3 of 3 stars, Deep challenge

Place one fair coin in each cell of a grid with 3 rows and 4 columns, and toss all 12 coins independently. A grid is successful if every row and every column contains an even number of heads. Zero heads counts as even.

(a) How many successful heads-and-tails patterns are there?

(b) What is the probability of success? Explain why satisfying the last row and last column does not impose two independent final restrictions.
1. Hint 1
  Choose the coins in the first two rows and first three columns freely.
2. Hint 2
  Use the last column to finish the first two rows, then the last row to finish the first three columns. What forces the bottom-right coin?
Answer

There are 64 successful patterns, with probability $1/64$ .

Full solution

First assign heads or tails freely to the 6 cells in the first two rows and first three columns.
There are $2^6=64$ assignments.
In each of those two rows, the fourth coin is now forced: choose it to make that row's number of heads even.

Next, in each of the first three columns, choose the bottom coin to make that column's heads count even.
Only the bottom-right coin remains.
Choose it so that the bottom row has an even number of heads.
We must check that this also makes the fourth column even.

All three rows are now even, so the total number of heads in the grid is even.
The first three columns each have an even number of heads, so their combined total is even.
Subtracting that total from the whole grid leaves an even number of heads in the fourth column.
Thus its condition is automatic.

Every initial choice produces exactly one successful grid, and every successful grid determines its initial 6 cells.
The count is therefore exactly 64.
All $2^{12}=4096$ full patterns are equally likely, so the probability is
$64/4096=1/64.$
The final row and column checks share the total-heads parity; treating all seven checks as independent would double-count one restriction.

Answer

There are 64 successful patterns, with probability $1/64$ .

Key idea

Choose independent information freely, then prove the remaining information is forced.
Problem 9: Mark complete
Problem 10 Stop at five

Difficulty: 3 of 3 stars, Deep challenge

A fair spinner has three equal sectors labeled 1, 2, and 3. Start with a total of 0. Spin repeatedly, adding each result to the total, and stop as soon as the total is at least 5. The spins are independent.

You win if the final total is exactly 5; you lose if it is greater than 5. Find the exact probability of winning. Is winning more likely, losing more likely, or are they equally likely? Give a method that accounts for games of different lengths.
1. Hint 1
  Work backward from totals close to 5 instead of drawing the entire tree of possible games.
2. Hint 2
  Let $p_k$ be the chance of hitting the target exactly when you still need $k$ points. The next spin leaves a smaller number of points to get, or overshoots.
Answer

Winning probability: $121/243$ . Losing probability: $122/243$ , so losing is slightly more likely.

Full solution

Let $p_k$ be the probability of reaching the target exactly when $k$ more points are needed.
Set $p_0=1$ : reaching the target is a win.
Overshooting is a loss, so use value 0 when the remaining amount is negative.

The next spin has three equally likely results.
For any positive $k$ , average the success probabilities after those results:
$p_k=(p_{k-1}+p_{k-2}+p_{k-3})/3.$
This equation is just a split into the three possible next spins; it does not assume that complete games have equal lengths or equal probabilities.

Work upward from the simplest cases.
We get $p_1=1/3$ , then $p_2=(1/3+1)/3=4/9$ , and
$p_3=(4/9+1/3+1)/3=16/27.$
Next,
$p_4=(16/27+4/9+1/3)/3=37/81.$
Finally,
$p_5=(37/81+16/27+4/9)/3=(37+48+36)/243=121/243.$

The starting total is 0, so 5 points are needed and the answer is $p_5$ .
Losing has probability $1-121/243=122/243$ , slightly larger.
Every game ends within five spins because each result adds at least 1; therefore there is no missing probability from a game that continues forever.

Answer

Winning probability: $121/243$ . Losing probability: $122/243$ , so losing is slightly more likely.

Key idea

For a process that repeats, solve smaller remaining tasks and combine their probabilities.
Problem 10: Mark complete