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Probability Puzzle

  • 13-11-2014 07:53PM
    #1
    ray giraffe
    Registered Users, Registered Users 2 Posts: 338 ✭✭


    You have 2 boxes: Box A and Box B.

    In Box A there are 2 gold coins and 1 silver coin.

    In Box B there are 2 silver coins and 1 gold coin.

    You choose a box at random, and then choose a coin at random from that box.

    Assuming that the coin you picked was gold, what is the probability that there is exactly 1 other gold coin in the box you picked?

    My first answer was [latex]\frac{1}{2}[/latex] but the correct answer is [latex]\frac{2}{3}[/latex] :o

    I wonder how could you reword the question to make the answer [latex]\frac{1}{2}[/latex]?


Comments

  • #2
    locum-motion
    Registered Users, Registered Users 2 Posts: 5,144 ✭✭✭locum-motion


    Explanation, as I understand it:

    Imagine that each coin has a number, 1,2,3 are in box A, and 4,5,6 are in Box B.

    There are 6 possible choices:
    1G
    2G
    3S
    4G
    5S
    6S

    If you have randomly picked a gold coin, you have chosen either 1, 2 or 4 (which is a 50:50 chance)

    Which of these options include "exactly 1 other gold coin in the box you picked"?

    1 does.
    2 does.
    4 does not.

    So, 2/3 have, and 1/3 has not.

    As to how you would reword the question:
    How about "Forget that coin. Randomly choose a box again. What are the chances that the new box has 3 coins in it?"


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