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If the head of a coin is worth 1 point and the tail is worth 2 points, [#permalink]
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Carcass wrote:
If the head of a coin is worth 1 point and the tail is worth 2 points, what is the probability that the sum of the results of 4 tosses is at least 5 points?

(A) 3/8
(B) 1/2
(C) 3/4
(D) 7/8
(E) 15/16

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Total outcomes = \(2^4 = 16\)
If the sum of the results is less than 5, we can have only 1 outcome for it i.e. HHHH = 4 points
So, we will have 15 cases where the sum of the results is 5 or more
Thus, required Prob. = \(\frac{15}{16}\)

Hence, option E
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