# 急：一個公正的錢幣丟八次

### 1 個解答

• 最佳解答

Conditional probablity: P(A / B) = P (A and B)/P(B)

P(2H/at least 1H occurs) = P([2H] and [at least 1H occurs]) / P(at least 1H occurs) = X/Y

X= P([2H] and [at least 1H occurs])= ?

So we know at least 1 H occurs, and there is another H somewhere.

How many ways can it be?

(1 H in 1st) (1 H in 2nd) (1 H in 3rd) (1H in 4th)

HHTTTTTT / THHTTTTT / TTHHTTTT / TTTHHTTT

HTHTTTTT / THTHTTTT / TTHTHTTT / TTTHTHTT

HTTHTTTT / THTTHTTT / TTHTTHTT / TTTHTTHT

HTTTHTTT / THTTTHTT / TTHTTTHT / TTTHTTTH

HTTTTHTT / THTTTTHT / TTHTTTTH

HTTTTTHT / THTTTTTH

HTTTTTTH

(1 H in 5th) (1 H in 6th) (1 H in 7th)

TTTTHHTT / TTTTTHHT / TTTTTTHH

TTTTHTHT / TTTTTHTH

TTTTHTTH

So the total ways is: 7 6 5 4 3 2 1=28

There are total of 2x2x2x2x2x2x2x2 = 256 ways

So X= P([2H] and [at least 1H occurs])= 28/256

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Y= P(at least 1H occurs) = 1- P(no H occurs) = 1- (1/256) = 255/256

Therefore,

P(2H/at least 1H occurs) = P([2H] and [at least 1H occurs]) / P(at least 1H occurs) = X/Y= (28/256)/(255/256) = 28/255 Q.E.D.

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