Question

in a certain population of people, 19% will contract disease a at some point in their lifetime, 20% will contract disease b at some point in their lifetime and 75% will not contract either of these diseases in their lifetime.
a person from this population is randomly chosen.
part (a) complete the probability table below. use two decimals in each of your answers.

part (b) find the probability that the person chosen contracts disease a or disease b. enter your answer to two decimals
part (c) find the probability that the person chosen contracts only one of these two diseases. (use two decimals)
part (d) suppose a person contracts disease a. what is the chance that this person will contract disease b? enter your answer to four decimals
part (e) if a person contracts disease b, what is the probability they will also contract disease a? (use four decimals)

Explanation:

Step1: Find the probability of contracting either disease

The probability of not contracting either disease \(P(A^{c}\cap B^{c}) = 0.75\). Using the formula \(P(A\cup B)=1 - P(A^{c}\cap B^{c})\), we have \(P(A\cup B)=1 - 0.75=0.25\).

Step2: Use the formula \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\)

Given \(P(A) = 0.19\), \(P(B)=0.20\), and \(P(A\cup B) = 0.25\). Substitute into the formula: \(0.25=0.19 + 0.20-P(A\cap B)\). Solving for \(P(A\cap B)\), we get \(P(A\cap B)=0.19 + 0.20 - 0.25=0.14\).

Step3: Find \(P(A\cap B^{c})\) and \(P(A^{c}\cap B)\)

\(P(A\cap B^{c})=P(A)-P(A\cap B)=0.19 - 0.14 = 0.05\).
\(P(A^{c}\cap B)=P(B)-P(A\cap B)=0.20 - 0.14=0.06\).
\(P(A^{c}\cap B^{c}) = 0.75\).

Part (b)

The probability that the person contracts Disease \(A\) or Disease \(B\) is \(P(A\cup B)\). As calculated above \(P(A\cup B)=0.25\).

Part (c)

The probability of contracting only one of the two diseases is \(P((A\cap B^{c})\cup(A^{c}\cap B))\). Since \((A\cap B^{c})\) and \((A^{c}\cap B)\) are mutually - exclusive, \(P((A\cap B^{c})\cup(A^{c}\cap B))=P(A\cap B^{c})+P(A^{c}\cap B)=0.05 + 0.06=0.11\).

Part (d)

Using the formula for conditional probability \(P(B|A)=\frac{P(A\cap B)}{P(A)}\). Substitute \(P(A\cap B) = 0.14\) and \(P(A)=0.19\), we get \(P(B|A)=\frac{0.14}{0.19}\approx0.7368\).

Part (e)

Using the formula for conditional probability \(P(A|B)=\frac{P(A\cap B)}{P(B)}\). Substitute \(P(A\cap B) = 0.14\) and \(P(B)=0.20\), we get \(P(A|B)=\frac{0.14}{0.20}=0.7000\).

Answer:

• Part (b): \(0.25\)
• Part (c): \(0.11\)
• Part (d): \(0.7368\)
• Part (e): \(0.7000\)