Looking for answers? Westonci.ca is your go-to Q&A platform, offering quick, trustworthy responses from a community of experts. Get quick and reliable answers to your questions from a dedicated community of professionals on our platform. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.
Sagot :
Let us call p the probability of occurence of the event D: "a student owns a dog"
and q the probability of occurence of the event C: "a student owns a cat"
In this exercice we have to assume that the events D and C are independent which means that none of them influence the occurence of the other. Because if they are not independent there really is noway to deduce q here.
In reality those events would probably not be independent since if I had a dog, I would really think twice before buying a cat ;)
Well if they are independent then you know that The Probability of occurence of D and C at the same time equals P(D)*P(C)=p*q
So we can deduce that 2/15=1/3*q
q=3*2/15
q=6/15
q=2/5
Once again this is theoritical
and q the probability of occurence of the event C: "a student owns a cat"
In this exercice we have to assume that the events D and C are independent which means that none of them influence the occurence of the other. Because if they are not independent there really is noway to deduce q here.
In reality those events would probably not be independent since if I had a dog, I would really think twice before buying a cat ;)
Well if they are independent then you know that The Probability of occurence of D and C at the same time equals P(D)*P(C)=p*q
So we can deduce that 2/15=1/3*q
q=3*2/15
q=6/15
q=2/5
Once again this is theoritical
Thank you for trusting us with your questions. We're here to help you find accurate answers quickly and efficiently. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.