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Three different coins are tossed simultaneously. Find the probability of getting :
(i) At least one head, (ii) At most two heads.
(i) At least one head, (ii) At most two heads.
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Total possible outcomes = $2^3 = 8$ (HHH, HHT, HTH, THH, HTT, THT, TTH, TTT)
(i) P (at least one head) = $1 - P(\text{no head}) = 1 - P(TTT) = 1 - \frac{1}{8} = \frac{7}{8}$
(ii) P (at most two heads) = $1 - P(\text{three heads}) = 1 - P(HHH) = 1 - \frac{1}{8} = \frac{7}{8}$
(i) P (at least one head) = $1 - P(\text{no head}) = 1 - P(TTT) = 1 - \frac{1}{8} = \frac{7}{8}$
(ii) P (at most two heads) = $1 - P(\text{three heads}) = 1 - P(HHH) = 1 - \frac{1}{8} = \frac{7}{8}$