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fiona asked in 科學及數學數學 · 1 decade ago

easy maths

1)A coin is thrown twice.Find the probability that

a)all are heads

b)at least one is head

c)exactly one is tail


Any explaination?

2 Answers

  • 1 decade ago
    Favorite Answer


    1st way of thinking:

    possible cases: HT, HH, TH, TT

    Pr[all heads] = correct case / possible case = 1/4

    2nd way of thinking (assume the two outcomes are independent):


    Pr[both head] = Pr[1st head] x Pr[2nd head] = (1/2)^2 = 1/4


    Pr[at least one head]

    = 1-Pr[no head]

    = 1-Pr[all tails]

    = 1-1/4 = 3/4


    Possible cases: HH, [HT], [TH], TT

    Pr[exactly one tail]

    =2/4 =1/2

    Source(s): myself
  • 1 decade ago

    a) 1/2 + 1/2 = 1/4

    b) 1/2 x 1/2 = 1/4

    c) 1/2

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