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Three Probability Problems

  1. Sep 12, 2008 #1
    Three Probability Problems urgent

    Q1.Prove that

    max[P(A),P(B)]<=P(A[tex]\cup[/tex]B)<=min[P(A)+P(B),1]


    Q2.A man is given n keys of which one fits the lock.He tries them successively without replacement.Find the probability that at the nth trial the lock will open.Also determine the expected numbers of trials.

    Q3.Thousand tickets are sold in a lottery in which there is one prize of Rs.500,four prizes of Rs.100 each & five prizes of Rs. 10 each.A ticket costs Rs.1.If X is your net gain when you buy one ticket,find E(X).
     
  2. jcsd
  3. Sep 12, 2008 #2

    statdad

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    Homework Helper

    Re: Three Probability Problems urgent

    General comments - if these don't help, show some work with your next post.

    1. What do you know about the comparisons of both [tex] A [/tex] and [tex] B [/tex] as events to the single event [tex] A \cup B [/tex]? this, together with a basic probability of probability of events, should give the first portion of the inequality. For the second portion of the inequality, think about 2 more things: the Addition rule and the maximum value of probability.

    2. If I understand the first question correctly, think about this: if you have 4 keys, and begin trying them, what can you say about the chance the door will be unlocked after you try all the keys? (I'm thinking I don't understand your question, because this idea seems too easy).
    For the expected value portion: pick one or two specific cases and work out the exact distribution (find the probability the first key alone works, that you need to try 2, that you need to try 3, etc.) and calculate the expected value. You should see a pattern to generalize. It's not the most direct approach, but may be the most informative

    3. You can write the percentages of tickets in each category: the net winnings are calculated based on face value minus cost of purchase. Set up the distribution (make sure you consider all cases and that the probabilities sum to 1) then calculate the expected value.

    Without some demonstrated calculations - that's all you get.
     
  4. Sep 12, 2008 #3
    Re: Three Probability Problems urgent

    Even I c'nt understand the first question.The question is found the same in a magazine.

    Plz help me.

    Thanks fr suggestion of 2nd & 3rd ques.
     
  5. Sep 12, 2008 #4

    statdad

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    Re: Three Probability Problems urgent

    The first question means this:
    Suppose you have two events [tex] A, B [/tex]. You need to show two things: first, that

    [tex]
    \max(\Pr(A), \Pr(B)) \le \Pr(A \cup B)
    [/tex]

    - this means that whichever of [tex] \Pr(A) [/tex] and [tex] \Pr(B) [/tex] is larger, it is still no bigger than [tex] \Pr(A \cup B) [/tex].

    Second, you need to show that

    [tex]
    \Pr(A \cup B) \le \max(\Pr(A) + \Pr(B), 1)
    [/tex]

    This statement tells you that [tex] \Pr(A \cup B ) [/tex] is always less-than-or-equal-to the larger of [tex] \Pr(A) + \Pr(B) [tex] and [tex] 1 [/tex].

    You do need to have some experience working with probability as well as the associated knowledge of its properties. If you haven't had a probability class you should look for a basic text: if you are in a class this is probably homework and you need to show, or explain, what work you've tried.
     
  6. Sep 12, 2008 #5

    statdad

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    Homework Helper

    Re: Three Probability Problems urgent

    sorry - missed a slash. My previous post should end
    the smaller of
    [tex] \Pr(A) + \Pr(B) [/tex] and [tex] 1 [/tex].
     
  7. Sep 12, 2008 #6
    Re: Three Probability Problems urgent




    3. The attempt at a solution


    We have P(A [tex]\cup[/tex] B)=P(A)+[P(b)-P(A[tex]\bigcap[/tex]B)]
    =P(B)+[P(A)-P(A[tex]\bigcap[/tex]B)]

    Then Can we say P(A[tex]\bigcup[/tex]B)[tex]\leq[/tex]max[P(A),P(B)] ?

    Plz help me.


    For the second part

    We have

    0[tex]\leq[/tex]P(A)[tex]\leq[/tex]1

    and

    0[tex]\leq[/tex]P(B)[tex]\leq[/tex]1


    Then min[P(A)+P(B),1]=P(A)+P(B).

    So we have to prove P(A[tex]\bigcup[/tex]B)[tex]\leq[/tex]P(A[tex]\bigcup[/tex]B)

    P(A)=P(A[tex]\bigcup[/tex]B)-[P(P(B)-P(A[tex]\bigcap[/tex]B)]

    and

    P(B)=P(A[tex]\bigcup[/tex]B)-[P(P(A)-P(A[tex]\bigcap[/tex]B)]

    Then,

    P(A)+P(B)=P(A[tex]\bigcup[/tex]B)+P(A[tex]\bigcap[/tex]B)

    Hence can we say

    P(A[tex]\bigcup[/tex]B)[tex]\leq[/tex]min[P(A)+P(B),1]

    Plz help me.
     
  8. Sep 13, 2008 #7
    Re: Three Probability Problems urgent

    3. The attempt at a solution for my 2nd question

    The man has n keys of which 1 is the right key.We are to find the probability that the lock will open at nth trail.

    So there are 1 right key for the lock and (n-1) other keys.

    Then total number of event points is (n-1)*(n-2)*(n-3)*............4*3*2*1


    The required event means that the right key is found at the nth trail(last trails) and (n-1) remaining keys are found at the preceding (n-1) trails.

    Then I cannot understand the no of favorable points.

    Please help me.
     
  9. Sep 13, 2008 #8
    Re: Three Probability Problems urgent

    Have you done inequalities? You know that [tex] \forall A \quad Pr(A) \geq 0[/tex] use that and also [tex] Pr(A)+Pr(B) \leq 2max(Pr(A),Pr(B))[/tex]. Remember that [tex] a-b \leq a[/tex] for [tex] b\geq0[/tex]
     
  10. Sep 13, 2008 #9
    Re: Three Probability Problems urgent


    Sorry I think my previous idea is wrong

    The Sample space will contain points

    n(n-1)(n-2)(n-3)....................{n-(n-1)}


    Since the lock opens at nth trails then the favorable cases are

    (n-1)(n-2)(n-3)..................{n-(n-1)}

    Then the required probability is 1/n.


    Am I right?

    Also see my attempt of the first question.
     
    Last edited: Sep 13, 2008
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