stat 213 assignment 1: problem 3 (1 point) if a and b are two mutually exclusive events with p(a)=0.4 and…

stat 213 assignment 1: problem 3 (1 point) if a and b are two mutually exclusive events with p(a)=0.4 and p(b)=0.5, find the following probabilities: a) p(a∩b)= b) p(a∪b)= c) p(a^c)= d) p(b^c)= e) p((a∪b)^c)= f) p(a∩b^c)=

stat 213 assignment 1: problem 3 (1 point) if a and b are two mutually exclusive events with p(a)=0.4 and p(b)=0.5, find the following probabilities: a) p(a∩b)= b) p(a∪b)= c) p(a^c)= d) p(b^c)= e) p((a∪b)^c)= f) p(a∩b^c)=

Answer

Explanation:

Step1: Recall definition of mutually - exclusive events

For mutually - exclusive events (A) and (B), (P(A\cap B)=0).

Step2: Use the addition rule for mutually - exclusive events

The formula for (P(A\cup B)) when (A) and (B) are mutually - exclusive is (P(A\cup B)=P(A)+P(B)). So (P(A\cup B)=0.4 + 0.5=0.9).

Step3: Use the complement rule

The complement rule states that (P(A^{c})=1 - P(A)). So (P(A^{c})=1 - 0.4 = 0.6).

Step4: Use the complement rule

The complement rule states that (P(B^{c})=1 - P(B)). So (P(B^{c})=1 - 0.5 = 0.5).

Step5: Use the complement rule

Since (P(A\cup B)=0.9), then (P((A\cup B)^{c})=1 - P(A\cup B)=1 - 0.9 = 0.1).

Step6: Since (A) and (B) are mutually - exclusive

(A\cap B^{c}=A), so (P(A\cap B^{c})=P(A)=0.4).

Answer:

a) (0) b) (0.9) c) (0.6) d) (0.5) e) (0.1) f) (0.4)