Let (A_{i}, i=1, ldots, 4), represent four events in a sample space, (S). For each of the

Question:

Let \(A_{i}, i=1, \ldots, 4\), represent four events in a sample space, \(S\). For each of the situations below, determine which assignment of probabilities are actually possible (i.e., do not contradict Kolmogorov's axioms), and which are not. Justify your answers.

a. \(P\left(A_{1}ight)=.3, P\left(A_{2}ight)=.3, P\left(A_{3}ight)=.2, P\left(A_{4}ight)=.2\)

b. \(P\left(A_{1}ight)=.3, P\left(A_{j}ight) \geq P\left(A_{1}ight)\) for \(j=2,3,4\)

c. \(P\left(A_{1}ight)=.7, P\left(A_{2}ight)=.6, P\left(A_{1} \cap A_{2}ight)=.1\)

d. \(P\left(A_{1}ight)=.7, P\left(A_{2}ight)=.6, P\left(A_{1} \cup A_{2}ight)=.1\)

e. \(P\left(A_{1}ight)=.3, P\left(A_{2}ight)=.4, P\left(A_{3}ight)=.1, P\left(A_{4}ight)=.2\) where \(A_{i} \subset A_{j}, i

f. \(P\left(A_{1}ight)=.4, P\left(A_{2}ight)=.3, P\left(A_{1} \cup A_{2}ight)=.5\) where \(A_{1} \cap A_{2}=\emptyset\)

Fantastic news! We've Found the answer you've been seeking!

Step by Step Answer:

Related Book For  book-img-for-question
Question Posted: