Question

d) what is the probability that a survivor was a male and in the 3rd class?
round final answer to 3 decimal places.

e) what is the probability that a survivor was a male or in the 3rd class?
round final answer to 3 decimal places.

Explanation:

Since no data (such as the number of survivors, number of male - 3rd class survivors, number of male survivors, number of 3rd - class survivors) is provided, we'll assume the following general approach using probability formulas. Let \(M\) be the event that a survivor is male and \(C_3\) be the event that a survivor is in the 3rd class.

d)

The probability of the intersection of two events \(P(M\cap C_3)\) is given by the formula \(P(M\cap C_3)=\frac{n(M\cap C_3)}{n(S)}\), where \(n(M\cap C_3)\) is the number of male survivors in the 3rd class and \(n(S)\) is the total number of survivors.

e)

The probability of the union of two events is given by the formula \(P(M\cup C_3)=P(M)+P(C_3)-P(M\cap C_3)=\frac{n(M)}{n(S)}+\frac{n(C_3)}{n(S)}-\frac{n(M\cap C_3)}{n(S)}\)

Step1: Recall probability formulas

For part d), use \(P(A\cap B)=\frac{n(A\cap B)}{n(S)}\). For part e), use \(P(A\cup B)=P(A) + P(B)-P(A\cap B)\)

Step2: Express in terms of counts

Let \(n(M)\) be the number of male survivors, \(n(C_3)\) be the number of 3rd - class survivors and \(n(M\cap C_3)\) be the number of male 3rd - class survivors and \(n(S)\) be the total number of survivors. So \(P(M)=\frac{n(M)}{n(S)}\), \(P(C_3)=\frac{n(C_3)}{n(S)}\) and \(P(M\cap C_3)=\frac{n(M\cap C_3)}{n(S)}\)

Since we have no actual values, we can't calculate a numerical answer. But if we had values for \(n(M)\), \(n(C_3)\), \(n(M\cap C_3)\) and \(n(S)\):

Answer:

d) Without data, unable to provide a numerical answer.
e) Without data, unable to provide a numerical answer.