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Someone please help me with a binomial distribution question watch

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    Show that, when two fair dice are thrown, the probability of obtaining a 'double' is 1/6, where a 'double' is defined as the same score on both dice. Four players play a board game which requires them to take it in turns to throw two fair dice. Each player throws the two dice once in each round. When a 'double' is thrown the player moves forward six squares. Otherwise the player moves forward one square. Findthe probability that a 'double' occurs exactly once in 4 of the first 5 rounds.
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    (Original post by richardhaha)
    show that, when two fair dice are thrown, the probability of obtaining a 'double' is 1/6, where a 'double' is defined as the same score on both dice. Four players play a board game which requires them to take it in turns to throw two fair dice. Each player throws the two dice once in each round. When a 'double' is thrown the player moves forward six squares. Otherwise the player moves forward one square. Findthe probability that a 'double' occurs exactly once in 4 of the first 5 rounds.
    x~b(4, 1/6)

    p(x=1)=>4*(1/6)*(5/6)^3=125/324

    y~b(5, 125/324)

    p(y=4)=>5*[1-(125/324)](125/324)^4=0.068
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    (Original post by PayneW)
    x~b(4, 1/6)

    p(x=1)=>4*(1/6)*(5/6)^3=125/324

    y~b(5, 125/324)

    p(y=4)=>5*[1-(125/324)](125/324)^4=0.068
    Thanks
 
 
 
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