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Math Help - Choosing letters with replacement.

  1. #1
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    Choosing letters with replacement.

    Five identical cards are labelled A, B, C, D, E. The cards are placed in a box and a card is selected at random. This card is replaced in the box, and a second random selection is made.

    a) Draw a grid to show the 25 possible outcomes.
    b) Given that neither card has a letter which appears in the word VICTORY, calculate the probability that both are vowels.

    I need help with b). What I did was consider it conditional probability. So I found the probability that both are vowels and don't appear in VICTORY which is 4/25 and I divided it by 1 becuase none of them appear in VICTORY. But my answer is wrong.

    I looked back at my work from the beginning of the semester and I divided by 16/25 instead, which was correct. But now I can't see why I would divide by 16/25.

    Does anyone know?
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  2. #2
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    Quote Originally Posted by bhuang View Post
    Five identical cards are labelled A, B, C, D, E. The cards are placed in a box and a card is selected at random. This card is replaced in the box, and a second random selection is made.

    a) Draw a grid to show the 25 possible outcomes.
    b) Given that neither card has a letter which appears in the word VICTORY, calculate the probability that both are vowels.

    I need help with b). What I did was consider it conditional probability. So I found the probability that both are vowels and don't appear in VICTORY which is 4/25 and I divided it by 1 becuase none of them appear in VICTORY. But my answer is wrong.

    I looked back at my work from the beginning of the semester and I divided by 16/25 instead, which was correct. But now I can't see why I would divide by 16/25.

    Does anyone know?
    Hi bhuang,

    C is the only letter of the 5 in VICTORY.

    Hence, A and E are two vowels of the 4 letters A, B, D, E that do not appear in VICTORY.

    Hence the probability that 2 cards do not appear in VICTORY is

    \frac{4}{5}\ \frac{4}{5}

    then the probability that the two cards are vowels is as you calculated.

    Given that the cards do not appear in VICTORY,
    there are 16 possibilities in choosing 2 cards.
    There are 4 ways to choose the vowels with replacement... AA, AE, EE, EA

    therefore there is a 1 in 4 chance the letters were vowels if the letters were not in VICTORY.

    Hence, in terms of conditional probability

    P(G|H)=\frac{P(G\cap H)}{P(H)}=\frac{\frac{4}{25}}{\frac{16}{25}}=\frac  {4}{25}\ \frac{25}{16}
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