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Conditional Probability

  1. Aug 9, 2012 #1
    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:)
     
  2. jcsd
  3. Aug 9, 2012 #2

    Ray Vickson

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    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
     
  4. Aug 9, 2012 #3
    Thank you! You have been very helpful
     
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