Question

assignment 1 replacement: problem 3 (1 point) if (p(a)=0.55, p(b)=0.45) and (p(acap b)=0.05), find the following probabilities: a) (p(acup b)=) b) (p(a^{c})=) c) (p(b^{c})=) d) (p(acap b^{c})=) e) (p(acap b)^{c}=)

Explanation:

Step1: Use the formula for $P(A\cup B)$

$P(A\cup B)=P(A)+P(B)-P(A\cap B)$
Substitute $P(A) = 0.55$, $P(B)=0.45$ and $P(A\cap B)=0.05$:
$P(A\cup B)=0.55 + 0.45- 0.05=0.95$

Step2: Use the formula for the complement probability $P(A^{c})$

$P(A^{c})=1 - P(A)$
Substitute $P(A)=0.55$:
$P(A^{c})=1 - 0.55 = 0.45$

Step3: Use the formula for the complement probability $P(B^{c})$

$P(B^{c})=1 - P(B)$
Substitute $P(B)=0.45$:
$P(B^{c})=1 - 0.45=0.55$

Step4: Use the formula $P(A\cap B^{c})=P(A)-P(A\cap B)$

Substitute $P(A)=0.55$ and $P(A\cap B)=0.05$:
$P(A\cap B^{c})=0.55 - 0.05=0.5$

Step5: Use the formula for the complement probability $P((A\cap B)^{c})$

$P((A\cap B)^{c})=1 - P(A\cap B)$
Substitute $P(A\cap B)=0.05$:
$P((A\cap B)^{c})=1 - 0.05 = 0.95$

Answer:

a) $0.95$
b) $0.45$
c) $0.55$
d) $0.5$
e) $0.95$