Lecture 2 - cont. Basic Probability

We start at [[handout01-ProbBasics-350F24.pdf#page=4]]:

(a) $P (A^{'}) = 1 - P (A) = .96$
$P (B^{'}) = 1 - P (B) = .67$

(b) $A \cap B$ are the set of submissions with both types of errors.

(c)

Use the diagram to get that it would be impossible to get

P (A \cap B)

so far. Namely, we could have

A

totally within

B

This would make $P (A \cap B) = P (A) = .14$ . That would be the maximum, as $A, B$ could be disjoint:

Then $P(A \cap B) = .0$ clearly.

Thus we have, with the current information that $0 \leq P (A \cap B) \leq .14$ .

(d) $A \cup B$ represents getting either a syntax error or a logic error (just any error for this case).

(e) $P (A \cup B) = P (A) + P (B) - P (A \cap B)$ (since we don't want to count the intersection bit twice). As a result then since $P (A \cap B) \in [0, .14]$ then we have:

P (A \cup B) = .14 + .33 - P (A \cap B) = .47 - P (A \cap B) \in [.33, .47]

This is the same addition rule we derived separately.
(f) If we are given the $P (A \cap B) = .05$ then we can use our formula to get:

P (A \cup B) = .14 + .33 - .05 = .47 - .05 = .42

(g) $P (A^{'} \cap B^{'}) = 1 - P (A \cup B) = 1 - .42 = .58$ .

(h) $P (A^{'} \cap B) = P (A \cup B) - P (A) = .42 - .14 = .28$ .

Contingency Table

We can make a contingency table for this situation:

	Syntax Errors $B$
Logic Errors $A$		.14
No Logic Errors $A^{'}$
	.33	1.0

With the supposition that $P (A \cap B) = .05$ :

	Syntax Errors $B$	No Syntax Errors $B^{'}$
Logic Errors $A$	.05	.09	.14
No Logic Errors $A^{'}$	.28	.58	.86
	.33	.67	1.0
The columns will add up to the bottom number, and the rows add to the last column.

(f) $P (A \cup B) = .14 + .33 - .05 = .42$ as we found before.
(g) $P (A^{'} \cap B^{'}) = .58$
(h) $P (A^{'} \cap B) = .28$

A 3 Example

(a)

(b) The

.07

probability seen in the diagram refers to

A \cap B^{'} \cap C^{'}

which is a purely minor defect.

#todo/school : Do Ch 1 #15, 27 📅 2024-09-25 ✅ 2024-09-24