Section 1.2

Basic Probability

Question 1.2.1

Suppose \(\Omega = \{1, 2, 3\}\), with \(\mathbb{P}(\{1\}) = \frac{1}{2}\), \(\mathbb{P}(\{2\}) = \frac{1}{3}\), and \(\mathbb{P}(\{3\}) = \frac{1}{6}\).

  1. What is \(\mathbb{P}(\{1, 2\})\)?

  2. What is \(\mathbb{P}(\{1, 2, 3\})\)?

By additivity of the probability measure: \[ \begin{aligned} \mathbb{P}(\{1,2\}) &= \mathbb{P}(\{1\}) + \mathbb{P}(\{2\}) \\ &= \frac{1}{2} + \frac{1}{3} \\ &= \frac{5}{6} \end{aligned} \] Therefore, \(\mathbb{P}(\{1,2\}) = \boldsymbol{\dfrac{5}{6}}\).

By additivity of the probability measure: \[ \begin{aligned} \mathbb{P}(\{1,2,3\}) &= \mathbb{P}(\{1\}) + \mathbb{P}(\{2\}) + \mathbb{P}(\{3\}) \\ &= \frac{1}{2} + \frac{1}{3} + \frac{1}{6} \\ &= 1 \end{aligned} \] This makes sense since \(\{1,2,3\} = S\), and we must always have \(\mathbb{P}(\Omega) = 1\). Therefore, \(\mathbb{P}(\{1,2,3\}) = \boldsymbol{1}\).

Question 1.2.2

Suppose \(\Omega = \{1, 2, 3\}\), with \(\mathbb{P}(\{1\}) = \frac{1}{2}\) and \(\mathbb{P}(\{1, 2\}) = \frac{2}{3}\). What must \(\mathbb{P}(\{2\})\) be?

Since \(\{1\}\) and \(\{2\}\) are disjoint and their union is \(\{1, 2\}\), additivity gives \[ \mathbb{P}(\{1,2\}) = \mathbb{P}(\{1\}) + \mathbb{P}(\{2\}) \] Rearranging, \[ \begin{aligned} \mathbb{P}(\{2\}) &= \mathbb{P}(\{1,2\}) - \mathbb{P}(\{1\}) \\ &= \frac{2}{3} - \frac{1}{2} \\ &= \frac{1}{6} \end{aligned} \] Therefore, \(\mathbb{P}(\{2\}) = \boldsymbol{\dfrac{1}{6}}\).

Question 1.2.3

Suppose \(\Omega = \{1, 2, 3\}\), and we try to define \(\mathbb{P}\) by \[ \mathbb{P}(\{1,2,3\}) = 1, \quad \mathbb{P}(\{1,2\}) = 0.7, \quad \mathbb{P}(\{1,3\}) = 0.5, \quad \mathbb{P}(\{2,3\}) = 0.7, \] \[ \mathbb{P}(\{1\}) = 0.2, \quad \mathbb{P}(\{2\}) = 0.5, \quad \mathbb{P}(\{3\}) = 0.3. \] Is \(\mathbb{P}\) a valid probability measure? Why or why not?

We must check if the probability axioms hold.

  1. \(\mathbb{P}(\Omega) = \mathbb{P}(\{1,2,3\}) = 1\) holds.

  2. \(\mathbb{P}(A) \ge 0\) for any \(A\) holds. All probabilities are non-negative.

  3. The probability of the union of two disjoint (\(\mathbb{P}(A \cup B)\)) events must equal the sum of their individual probabilities(\(\mathbb{P}(A) + \mathbb{P}(B)\)). In particular, since \(\{2\}\) and \(\{3\}\) are disjoint with union \(\{2,3\}\), we would need \[ \mathbb{P}(\{2,3\}) = \mathbb{P}(\{2\}) + \mathbb{P}(\{3\}) \] Checking this: \[ \mathbb{P}(\{2\}) + \mathbb{P}(\{3\}) = 0.5 + 0.3 = 0.8 \neq 0.7 = \mathbb{P}(\{2,3\}) \] Since additivity fails for this pair of disjoint events, \(\mathbb{P}\) is not a valid probability measure.

Question 1.2.4

Suppose \(\Omega = \{1, 2, 3\}\), and \(\mathbb{P}(\{1\}) = \mathbb{P}(\{3\}) = 2\mathbb{P}(\{2\})\). Compute \(\mathbb{P}(\{1\})\), \(\mathbb{P}(\{2\})\), and \(\mathbb{P}(\{3\})\).

Since \(\Omega = \{1,2,3\}\), we must have \(\mathbb{P}(\Omega) = 1\), i.e., \[ 1 = \mathbb{P}(\{1\}) + \mathbb{P}(\{2\}) + \mathbb{P}(\{3\}) \] Using the given relationships \(\mathbb{P}(\{1\}) = 2\mathbb{P}(\{2\})\) and \(\mathbb{P}(\{3\}) = 2\mathbb{P}(\{2\})\), this becomes \[ \begin{aligned} 1 &= 2\mathbb{P}(\{2\}) + \mathbb{P}(\{2\}) + 2\mathbb{P}(\{2\}) \\ &= 5\,\mathbb{P}(\{2\}) \end{aligned} \] so \[ \mathbb{P}(\{2\}) = \frac{1}{5} \] Then, \[ \mathbb{P}(\{1\}) = 2\mathbb{P}(\{2\}) = \frac{2}{5}, \qquad \mathbb{P}(\{3\}) = \mathbb{P}(\{1\}) = \frac{2}{5} \] Therefore, \(\mathbb{P}(\{1\}) = \boldsymbol{\dfrac{2}{5}}\), \(\mathbb{P}(\{2\}) = \boldsymbol{\dfrac{1}{5}}\), and \(\mathbb{P}(\{3\}) = \boldsymbol{\dfrac{2}{5}}\).

Question 1.2.5

A quality control inspector examines items coming off a production line and classifies each as defective (D) or non-defective (N). The inspector examines items one at a time and stops as soon as they find 2 defective items, or after examining 4 items, whichever comes first.

  1. Write out the sample space \(\Omega\).

\(\Omega = \{DD,NDD,DND,NNDD,NDND,DNND,NNNN,DNNN,NDNN,NNDN, NNND\}\)

  1. Let \(A\) be the event that the inspector stops after exactly 3 items. List the outcomes in \(A\)

If the inspector stops after three items, it means they found two defects, and these two defects were not the first two. \(A = \{NDD, DND\}\).

  1. Let \(B\) be the event that fewer than 2 defective items are found in total. List the outcomes in \(B\).

If there are fewer than two defective items in total, then that means there are either 0 or 1 defectives across the whole sequence. And that means we’d need four items.

\(B = \{NNNN, DNNN, NDNN, NNDN, NNND\}\)

  1. What is \(A \cap B\)? Describe this subset in words.

Sequences where the inspector stopped after three items and there were fewer than 2 defects. This isn’t possible as an inspector will only stop at 3 items if there are 2 defects. Therefore, \(A \cap B = \emptyset\)