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Math Help - Multiple Choice Exam & Jelly Beans in a Bag

  1. #1
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    Multiple Choice Exam & Jelly Beans in a Bag

    Probability drives me crazy
    I couldn't figure out how it solves
    could you help me please

    Multiple Choice Exam
    Each question on a 5 questions multiple choice examination has 4 choices,only one of which is correct . If a student answers each question on a random fashion, what's the probability that the student answers exactly 2 question incorrectly ?
    My answer :
    # (S)=1024 (4^5)
    # (E)=10 (5C3) I said if the student answers 2 question incorrectly is the same as if he answers ignore any 2 question
    P(E)=  10/1024

    tell me how to solve it plz

    BUT the right answer is ( 45/512)

    Jelly Beans in a Bag

    Abag contains 4 red and 6 green jelly beans.
    a) if 2 jelly beans are randomly selected in succession with replacement, determine the probability that both are red.
    b)if the selection is made without replacement .determine the probability that both are red.

    My answer
    #( S)=24
    a)# (E)= 4c2<br />
P(E)= 6/24=1/4<br />
    b)#( E)= 4p2<br />
P(E)= 12/24=1/2<br />

    BUT the right answer is

    a)4/25<br />

    <br />
b)2/15<br />
    Tell me please how to solve them


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  2. #2
    MHF Contributor ebaines's Avatar
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    The first one is solved by realizing that the student gets 3 correct and two incorrect, that the probability of a correct answer on any one question is 1/4, and the probablilty of an incorrect answer is 3/4. The right and wrong answers can be arranged in any one of 5C2 ways. So you get:

    5C2*(1/4)^3 * (3/4)^2 = 10*9/1024 = 45/512.

    For the second problem: there are 4 reds and 10 total, so the probability of pulling a red on the first draw is 4/10. If you put that red one back and draw again, the probablity of drawing red a second time is also 4/10. But if you don't replace it, then for the second draw there are only 3 reds out of the 9 that are left, so the probability of a second red is 3/9. Putting it together:
    a) prob of two reds in a row with replacement = 4/10 * 4/10 = 4/25.
    b) prob of two reds in a row without replacement = 4/10 * 3/9 = 2/15.
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