Counting and chance in Python

Fill in the blank

What would Python display? Indexing

What would Python display?

>>> v = [3, 1, 4, 1, 5, 9]
>>> v[1]

Your answer

Use an integer, decimal, or fraction such as 1/2. An exact value is expected.

Hints

1

Python numbers positions from 0, not from 1.

2

v[0] is 3. Which entry comes next?

Reference solution

1. Python positions start at 0, so v[1] is the second entry. In R, v[1] would be the first entry, 3.

Source: Adapted from Blitzstein & Hwang, Introduction to Probability, §1.8 (R), rewritten in Python · Original exercise; the book's R examples are paraphrased, not copied

Single choice

What would Python display? No recycling

What happens here?

>>> v = [3, 1, 4]
>>> v + 3

Choose one answer.

[6, 4, 7]

[3, 1, 4, 3]

A TypeError

Hints

1

+ between two lists joins them. Is 3 a list?

2

Python does not repeat a shorter value to fit a longer one.

Reference solution

Python raises TypeError: can only concatenate list (not "int") to list. For lists, + means concatenation, and it needs a list on both sides. Write [x + 3 for x in v] to add 3 to each entry, which gives [6, 4, 7].

Source: Adapted from Blitzstein & Hwang, Introduction to Probability, §1.8 (R), rewritten in Python · Original exercise; the book's R examples are paraphrased, not copied

Python

Binomial coefficients by recursion

Complete choose(n, k) recursively, using Pascal’s rule $\binom{n}{k}=\binom{n-1}{k-1}+\binom{n-1}{k}$ . Do not use math.comb or factorials. Return 0 when $k<0$ or $k>n$ .

Your Python code

Python function exercise. Tab moves focus; use the indent buttons for selected lines.

Public tests

1. Small case

>>> choose(5, 2)
10

2. Middle of a row

>>> choose(10, 5)
252

3. Choose none or all

>>> choose(4, 0)
1
>>> choose(4, 4)
1

4. Too many

>>> choose(3, 5)
0

Output and test details

Hints

1

There are two base cases: impossible choices, and choosing none or all.

2

If $k$ is 0 or equal to $n$ , there is exactly one way.

3

Otherwise, return the sum of two smaller choose calls.

Reference solution

def choose(n, k):
    if k < 0 or k > n:
        return 0
    if k == 0 or k == n:
        return 1
    return choose(n - 1, k - 1) + choose(n - 1, k)

The first base case handles impossible requests; the second is the edge of Pascal’s triangle. Each recursive call lowers $n$ , so the recursion ends. It makes many repeated calls, which is fine for small $n$ and a good reason to meet memoization later.

Source: Adapted from Blitzstein & Hwang, Introduction to Probability, §1.8 (R), rewritten in Python · Original exercise; the book's R examples are paraphrased, not copied

Fill in the blank

What would Python display? Adding truth values

What would Python display?

>>> deck = [2, 1, 3, 4]
>>> sum([card == position for position, card in enumerate(deck, start=1)])

Your answer

Use an integer, decimal, or fraction such as 1/2. An exact value is expected.

Hints

1

The list inside is [False, False, True, True].

2

In arithmetic, True counts as 1 and False as 0.

Reference solution

2. Cards 3 and 4 are in their own positions, so the list is [False, False, True, True], and its sum counts the True values.

Source: Adapted from Blitzstein & Hwang, Introduction to Probability, §1.8 (R), rewritten in Python · Original exercise; the book's R examples are paraphrased, not copied

Python

Count the matches

Complete count_fixed_points(deck). deck lists the cards in the order they are turned over; card $i$ matches when it is turned over $i$ th. Return the number of matches.

Your Python code

Python function exercise. Tab moves focus; use the indent buttons for selected lines.

Public tests

1. Every card matches

>>> count_fixed_points([1, 2, 3])
3

2. One match

>>> count_fixed_points([2, 1, 3])
1

3. No matches

>>> count_fixed_points([3, 1, 2])
0
>>> count_fixed_points([2, 1])
0

4. Empty deck

>>> count_fixed_points([])
0

Output and test details

Hints

1

enumerate(deck, start=1) gives each position together with its card.

2

Compare each card with its position, then add up the results.

Reference solution

def count_fixed_points(deck):
    return sum(card == position for position, card in enumerate(deck, start=1))

Each comparison is True or False, which sum counts as 1 or 0. An empty deck has no matches, so the sum is 0.

Source: Adapted from Blitzstein & Hwang, Introduction to Probability, §1.8 (R), rewritten in Python · Original exercise; the book's R examples are paraphrased, not copied

Python

How many people are needed?

birthday_match(k, days) is provided. Complete people_needed(p, days): return the smallest number of people $k\ge1$ for which birthday_match(k, days) is at least p. This is what R’s qbirthday computes. Assume $0<p\le1$ .

Your Python code

Python function exercise. Tab moves focus; use the indent buttons for selected lines.

Public tests

1. Even odds

>>> people_needed(0.5)
23

2. Ninety percent

>>> people_needed(0.9)
41

3. Ninety-nine percent

>>> people_needed(0.99)
57

4. A shorter year

>>> people_needed(0.5, days=2)
2

Output and test details

Hints

1

Start with one person and add people until the probability is large enough.

2

A while loop fits: keep going while birthday_match(k, days) < p.

Reference solution

def people_needed(p, days=365):
    k = 1
    while birthday_match(k, days) < p:
        k += 1
    return k

The loop always stops: with days + 1 people a shared birthday is certain, so the probability reaches 1.

Source: Adapted from Blitzstein & Hwang, Introduction to Probability, §1.8 (R), rewritten in Python · Original exercise; the book's R examples are paraphrased, not copied