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  • November 6, 2021
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STA1000S 2021
Week 1 Weekly Exercises – Set Theory & Basic Probability

*Solutions*
Question 1:

1000 applicants have applied for a coding job at a data analytics company known as Data
Wizard. The applicants are proficient in different coding languages, but the company prefers
applicants who can code in 𝑅𝑅, 𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩 or 𝑪𝑪++. We will use a three-variable Venn Diagram to
represent the problem. Let the top circle indicate the programmers who are proficient at
coding in 𝑅𝑅 (Denote this set with an R). Let the left circle indicate the programmers who are
proficient at coding in 𝐩𝐩𝐲𝐲𝐭𝐭𝐭𝐭𝐭𝐭𝐭𝐭 (Denote this set with a P). Finally, let the right circle indicate
the programmers who are proficient at coding in 𝑪𝑪++ (Denote this set with a C). The HR
department has determined the following information from the applicants:

• 40 applicants are proficient in all three coding languages.
• The total number of applicants that are proficient in 𝑅𝑅 = 595.
• The total number of applicants that are proficient in 𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩 = 380.
• The total number of applicants that are proficient in 𝑪𝑪++ = 230.
• 89 applicants that applied were proficient in none of the three coding
languages.
• 121 applicants were proficient in both 𝑅𝑅 and 𝑪𝑪++.
• 74 applicants were proficient in 𝑅𝑅 and 𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩 but not in C++.

a) Determine how many applicants are proficient at coding in 𝑅𝑅 but not in 𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩 and
not in 𝑪𝑪++ (i.e., Determine 𝑛𝑛(𝑅𝑅 ∩ 𝑃𝑃� ∩ 𝐶𝐶̅ ) ).

We know that 𝑛𝑛(𝑅𝑅 ∩ 𝑃𝑃 ∩ 𝐶𝐶̅ ) = 74 (i.e., 74 applicants were proficient in 𝑅𝑅 and
𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩 but not in 𝑪𝑪++). We are also told that 𝑛𝑛(𝑅𝑅 ∩ 𝐶𝐶) = 121. We can then
determine the following:

𝑛𝑛(𝑅𝑅 ∩ 𝐶𝐶 ∩ 𝑃𝑃� ) = 𝑛𝑛(𝑅𝑅 ∩ 𝐶𝐶) − 𝑛𝑛(𝑅𝑅 ∩ 𝑃𝑃 ∩ 𝐶𝐶)
= 121 − 40 = 81.

Therefore, 81 applicants are proficient in 𝑅𝑅 and 𝑪𝑪++ but not in 𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩. We can now
use the total number of applicants that are proficient in 𝑅𝑅 (595 applicants) to
determine the answer to this question. This is carried out as follows:

∴ 𝑛𝑛(𝑅𝑅 ∩ 𝑃𝑃� ∩ 𝐶𝐶̅ ) = 𝑛𝑛(𝑅𝑅) − 𝑛𝑛(𝑅𝑅 ∩ 𝑃𝑃 ∩ 𝐶𝐶) − 𝑛𝑛(𝑅𝑅 ∩ 𝐶𝐶 ∩ 𝑃𝑃�) − 𝑛𝑛(𝑅𝑅 ∩ 𝑃𝑃 ∩ 𝐶𝐶̅ )
= 595 − 40 − 81 − 74 = 400.

,b) Determine how many applicants are proficient at coding in 𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩 and 𝑪𝑪++ but not
�) ).
in 𝑅𝑅 (i.e., Determine 𝑛𝑛(𝑃𝑃 ∩ C ∩ R

Let 𝑛𝑛(𝑃𝑃 ∩ 𝑅𝑅� ∩ 𝐶𝐶̅ ) = 𝑎𝑎, 𝑛𝑛(𝑃𝑃 ∩ 𝐶𝐶 ∩ 𝑅𝑅� ) = 𝑏𝑏 and 𝑛𝑛(𝐶𝐶 ∩ 𝑅𝑅� ∩ 𝑃𝑃�) = 𝑐𝑐. We can then set
up the following simultaneous equations:

① 𝑎𝑎 + 𝑏𝑏 + 40 + 74 = 380
② 𝑏𝑏 + 𝑐𝑐 + 40 + 81 = 230
③ 𝑎𝑎 + 𝑏𝑏 + 𝑐𝑐 + 74 + 40 + 81 + 400 = 1000 − 89 = 911

The above equations are then simplified as follows:

① 𝑎𝑎 + 𝑏𝑏 = 266
② 𝑏𝑏 + 𝑐𝑐 = 109
③ 𝑎𝑎 + 𝑏𝑏 + 𝑐𝑐 = 316

We can sub ① into ③:

∴ 𝑎𝑎 + 𝑏𝑏 + 𝑐𝑐 = 266 + 𝑐𝑐 = 316.
⇒ 𝑐𝑐 = 50 ④

Sub ④ into ②:

∴ 𝑏𝑏 + 𝑐𝑐 = 𝑏𝑏 + 50 = 109.
⇒ 𝑏𝑏 = 59

We therefore know that 𝑛𝑛(𝑃𝑃 ∩ 𝐶𝐶 ∩ 𝑅𝑅� ) = 59.

c) Suppose that applicants are randomly chosen. What is the probability that a
randomly chosen applicant would either be proficient at coding in 𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩 but not in
𝑪𝑪++ and not in 𝑅𝑅 OR proficient at coding in 𝑪𝑪++ but not in 𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩 and not in 𝑅𝑅
(Hint: First determine 𝑛𝑛(𝑃𝑃 ∩ 𝑅𝑅� ∩ 𝐶𝐶̅ ) and 𝑛𝑛(𝐶𝐶 ∩ 𝑅𝑅� ∩ 𝑃𝑃�) ).

Using the set of simultaneous equations from the previous question, we can
determine 𝑛𝑛(𝑃𝑃 ∩ 𝑅𝑅� ∩ 𝐶𝐶̅ ). Note that we have already determined that
𝑛𝑛(𝐶𝐶 ∩ 𝑅𝑅� ∩ 𝑃𝑃� ) = 50 in the previous question. Using ①:

∴ 𝑎𝑎 + 𝑏𝑏 = 𝑎𝑎 + 59 = 266.
⇒ 𝑎𝑎 = 𝑛𝑛(𝑃𝑃 ∩ 𝑅𝑅� ∩ 𝐶𝐶̅ ) = 207.

, Therefore, probability that a randomly chosen applicant would either be proficient at
coding in 𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩 but not in 𝑪𝑪++ and not in 𝑅𝑅 OR proficient at coding in 𝑪𝑪++ but not
in 𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩 and not in 𝑅𝑅:

𝑛𝑛(𝑃𝑃 ∩ 𝑅𝑅� ∩ 𝐶𝐶̅ ) 𝑛𝑛(𝐶𝐶 ∩ 𝑅𝑅� ∩ 𝑃𝑃�)
Required probability = +
1000 1000
207 50 257
= + = = 0.257.
1000 1000 1000
Question 2:

Given that Pr(𝐴𝐴) = 0.45, Pr(𝐵𝐵) = 0.39 and Pr(𝐴𝐴 ∩ 𝐵𝐵) = 0.15, use the Venn Diagram below to
calculate the following probabilities:

a) Pr(𝐴𝐴̅ ∩ 𝐵𝐵� )
Pr(𝐴𝐴̅ ∩ 𝐵𝐵� ) = 1 − Pr(𝐴𝐴 ∪ 𝐵𝐵) = 1 − 0.69 = 0.31. The region of interest is shown
using the following diagram:




b) Pr�𝐵𝐵� ∩ (𝐴𝐴 ∪ 𝐵𝐵)�

Pr�𝐵𝐵� ∩ (𝐴𝐴 ∪ 𝐵𝐵)� = Pr(𝐵𝐵� ∩ 𝐴𝐴) = 0.3. The region of interest is shown using the
following diagram:

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