Calculating Probabilities Using Probability Rules

Can someone please review my work and let me know if I'm doing them right? If not, please let me know how I can fix it. Thank you.



A, B, and C are events of a probability experiment. Given that
P(A)=0.25, P(B)=0.75, P(C)=.35
determine the following:


a. P(A or B) if P(A and B) = 0.27

P(A or B) = P(A) + P(B) - P(A and B)
= .25 + .75 - .27
= .73



b. P(A or C) if A and C are mutually exclusive

P(A) + P(C)
P(.25) + P(.35)
= .6



c. P(B and C) if P(B or C) = 0.15

.15 - .75 + .35
= -.4



d. P(C[sup:ryo85qwa]c[/sup:ryo85qwa])

P(C[sup:ryo85qwa]c[/sup:ryo85qwa]) = 1 - P(C)= 1 - .35
= .65



e. P(A and C) if A and C are independent.

P(A) . P(C)
P(.25) . P(.35)
= .088

Hello, boydoineedhelp!

Only part (c) is off . . . Good work!

\(\displaystyle A, B,\text{ and }C\text{ are events of a probability experiment.}\)

\(\displaystyle \text{Given: }\;P(A)\,=\,0.25,\;\;P(B)\,=\,0.75,\;\; P(C)\,=\,0.35\)

\(\displaystyle \text{Find: }\;(c)\;P(B \cap C)\:\text{ if }P(B \cup C) \,=\, 0.15\)

Click to expand...

\(\displaystyle \text{Use the formula: }\;\underbrace{P(B \cup C)}_{0.15} \;=\; \underbrace{P(B)}_{0.75} + \underbrace{P(C)}_{0.35} - P(B \cap C)\)

\(\displaystyle \text{We have: }\:0.15 \;=\;0.75 + 0.35 - P(B \cap C)\)


\(\displaystyle \text{Therefore: }\(B \cap C) \:=\:0.75 + 0.35 - 0.15 \:=\:0.95\)

Thanks! I knew that question C was definitely wrong because it was a negative. I actually figured out the same answer (.95) you got late last night. I wasn't sure if .95 was right since the wording for the problem was different and I was not sure how to show the breakdown of the formula. Now I see how it comes out to that. Thanks again. I appreciate it.