Conditional Probability

In summary, the conversation discusses a question regarding probability and explores whether two events are independent or not. It is determined that the events are not independent and a solution is provided for one part of the question.
  • #1
porroadventum
34
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1. I am going over some past Probability exam papers and cannot solve this question. Any help or advice would be much appreciated!


2. David eats cereal for lunch 60% of days. If he had ice cream for breakfast, then the probability that he eats cereal for lunch is only 0.25. If he didn't eat ice cream for breakfast then he would not eat cereal for lunch with probability 0.3.

(a) Are the events that David ate cereal for lunch and ice cream for breakfast independent?
(b) Show that P(David has ice cream for breakfast)=2/9



3. I (think I) have been able to work out part (a): Let the event cereal for lunch be A and ice cream for breakfast be B. The events A and B are independant if P(A[itex]\cap[/itex]B)= P(a)* P(B). So P(A given B)*P(B)= P(A)* P(B) which gives us P(A given B)= P(A) and then when we sub in the values given in the question we get 0.25= 0.6 which is not true and thus proves they are not independent. Is this correct?

For part b I have been using the partition theorem to try to show that P(B)=2/9. So P(B)= P(B given A)*P(A)+ P(B given Ac)*P(ac) which gives me P(B)*(0.75)=. And this is as far as I can get because I cannot work out how to find P(B[itex]\cap[/itex]Ac)?? More than likely I have gone completely wrong from the very beginning. Any help would be very much appreciated please!

Thank you:)
 
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  • #2
porroadventum said:
1. I am going over some past Probability exam papers and cannot solve this question. Any help or advice would be much appreciated!


2. David eats cereal for lunch 60% of days. If he had ice cream for breakfast, then the probability that he eats cereal for lunch is only 0.25. If he didn't eat ice cream for breakfast then he would not eat cereal for lunch with probability 0.3.

(a) Are the events that David ate cereal for lunch and ice cream for breakfast independent?
(b) Show that P(David has ice cream for breakfast)=2/9



3. I (think I) have been able to work out part (a): Let the event cereal for lunch be A and ice cream for breakfast be B. The events A and B are independant if P(A[itex]\cap[/itex]B)= P(a)* P(B). So P(A given B)*P(B)= P(A)* P(B) which gives us P(A given B)= P(A) and then when we sub in the values given in the question we get 0.25= 0.6 which is not true and thus proves they are not independent. Is this correct?

For part b I have been using the partition theorem to try to show that P(B)=2/9. So P(B)= P(B given A)*P(A)+ P(B given Ac)*P(ac) which gives me P(B)*(0.75)=. And this is as far as I can get because I cannot work out how to find P(B[itex]\cap[/itex]Ac)?? More than likely I have gone completely wrong from the very beginning. Any help would be very much appreciated please!

Thank you:)

I prefer to use a notation where the meaning of the symbols is apparent at once, so let I = {ice cream for breakfast} and C = {cereal for lunch}. You are given P(C) = 0.6 = 3/5, P(C|I) = 0.25 = 1/4 and P(Cc|Ic) = 0.3 = 3/10. Thus, P(C|Ic) = 7/10. Since P(C|I) and P(C|Ic) are different, C and I are dependent, as you said.

P(C) = P(C & I) + P(C & Ic) = P(C|I)P(I) + P(C|Ic)P(Ic), and P(Ic) = 1-P(I). You can solve for P(I).

RGV
 
  • #3
Thank you! You have been very helpful
 

What is conditional probability?

Conditional probability is a measure of the probability of an event occurring, given that another event has already occurred. It can be represented mathematically as P(A|B), where A is the event of interest and B is the event that has already occurred.

How is conditional probability calculated?

The formula for calculating conditional probability is P(A|B) = P(A∩B) / P(B), where P(A∩B) represents the probability of both events A and B occurring together, and P(B) represents the probability of event B occurring.

What is the relationship between conditional probability and independence?

If two events A and B are independent, then the conditional probability of A given B is equal to the unconditional probability of A. In other words, the occurrence of B does not affect the probability of A occurring. However, if two events are not independent, then the conditional probability of A given B will be different from the unconditional probability of A.

How is conditional probability used in real life?

Conditional probability is used in many fields, such as finance, medicine, and marketing. In finance, it can be used to calculate the likelihood of certain market trends based on past data. In medicine, it can help doctors make more accurate diagnoses by taking into account the patient's symptoms and medical history. In marketing, it can be used to predict consumer behavior and target specific demographics.

What is the difference between conditional probability and joint probability?

Conditional probability is the probability of an event occurring given that another event has already occurred. Joint probability, on the other hand, is the probability of two or more events occurring together. While conditional probability focuses on the relationship between two events, joint probability considers the probability of multiple events occurring simultaneously.

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