Statistics with Probability

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  • #12966

    Ezra Halleck
    Participant

    4. Suppose you roll two fair dice. Let A= doubles, B= sum is 6
    (a) Are A and B disjoint?
    (b) Determine P(A), P(B) and P(A and B).
    (c) Are A and B independent?

    #16140

    dannyCR09
    Member

    4.
    a) A and B are disjoint events.

    b) P(A) = 6/36 or 1/6
    P(B) = 2/36 or 1/18
    P(A & B) = 8/36 or 2/9

    c) No, A and B are not independent.
    P(A ∩ B) = P(A) * P(B)
    0 = 6/36 * 2/36
    0 = 1/108
    X

    #16141

    Ezra Halleck
    Participant

    I suggest that you start off by listing exactly what A and B are. Then you can perhaps begin answering the questions. To be blunt, you write P(A & B) = 8/36 or 2/9 but earlier, you said that they were disjoint. It would be helpful to have definitions, since we have not seen this material for some time.

    #16142

    dannyCR09
    Member

    When you say P(A and B), do you mean the union or the intersection?

    #16143

    Ezra Halleck
    Participant

    and=intersection
    or=union

    #16144

    dannyCR09
    Member

    Roll of two fair dice
    P(A) = doubles , P(B) = even sum of 6

    4.
    a) A and B are disjoint events.
    Disjoint – when two events cannot occur at the same time

    b) The probability of P(A) happening – getting a double when you roll the two dice (1,1; 2,2; 3,3, etc.) = 6/36 or 1/6
    The probability ofP(B) happening – getting two even numbers to add up to 6 (4,2; 2,4) = 2/36 or 1/18
    The intersection of P(A & B) is 0 because A and B cannot occur at the same time.

    c) A is independent of B if P(A ∩ B) = P(A) * P(B) and vice versa
    No, A and B are not independent.
    P(A ∩ B) = P(A) * P(B)
    0 = 6/36 * 2/36
    0 = 1/108
    X

    #16154

    Ezra Halleck
    Participant

    I see where the problem is now. There are more ways to get a sum of 6, e.g., 1,5.

    #16155

    dannyCR09
    Member

    So, when u put “B= even sum is 6”, did you mean even numbers that sum up to six or just any numbers that sum up to 6?

    #16187

    Corey
    Member

    DIsjoint events means that that are independent of each other correct?

    #16197

    Ezra Halleck
    Participant

    Danny wrote:
    “Disjoint – when two events cannot occur at the same time.”
    This is not quite it. Remember, an event is a collection of outcomes. Disjoint means no intersection. So the outcomes in one event are not part of the other event and vice verse.

    #16206

    Afzal
    Participant

    a. I still didn’t get it professor.

    b. Doubles,P(A)={ (1,1),(2,2),(3,3),(4,4),(5,5),(6,6)}=6/36=1/6
    sum is 6, P(B)={ (1,5),(2,4),(3,3),(4,2),(5,1)}=5/36
    P(A and B)={(3,3)}=1/36

    c. P(A and B) = P(A) * P(B)
    1/36=1/6*5/36
    1/36=11/36 which is not true ,so it is not independent.

    #16209

    Ezra Halleck
    Participant

    To answer a, the 2 events in question have a common outcome, namely (3,3), so the events are not disjoint.

    #16224

    soumanusa
    Member

    4.
    a) A and B are disjoint events.

    b) P(A) = 6/36 or 1/6
    P(B) = 2/36 or 1/18
    P(A & B) = 8/36 or 2/9

    c) No, A and B are not independent.
    P(A ∩ B) = P(A) * P(B)
    0 = 6/36 * 2/36
    0 = 1/108
    X

    #16225

    Afzal
    Participant

    hi soumanusa,
    how you get P(B)=2/36?. P(B) is sum of 6. and i think { (1,5),(2,4),(3,3),(4,2),(5,1)} all set comply with the condition. so i think P(B) will be 5/36.

    #16232

    Ezra Halleck
    Participant

    Afzal is correct and in fact just 2 posts before your post, soumanusa, he has a correct solution except for a, which I have answered in the post right before yours.

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