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Question

# A coin is tossed three times then find the probability of (i) getting head on middle coin. (ii) getting exactly one tail. (iii) getting no tail.

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Solution

## Given: A coin is tossed three times. ∴ Sample space = {HHH, THH, HTH, HHT, THT, TTH, HTT, TTT} Or, n(S) = 8 (i) Let X be the event of getting head on the middle coin. X = {HHH, THH, HHT, THT} ∴ n(X) = 4 Thus, we have: P(X) = $\frac{n\left(\mathrm{X}\right)}{n\left(\mathrm{S}\right)}$ $⇒$P(X) =$\frac{4}{8}=\frac{1}{2}$ (ii) Let Y be the event of getting exactly one tail. Y = {THH, HTH, HHT} ∴ n(Y) = 3 Thus, we have: P(Y) = $\frac{n\left(\mathrm{Y}\right)}{n\left(\mathrm{S}\right)}$ $⇒$P(Y) = $\frac{3}{8}$ (iii) Let Z be the event of getting no tail. Z = {HHH} ∴ n(Z) = 1 Thus, we have: P(Z) = $\frac{n\left(\mathrm{Z}\right)}{n\left(\mathrm{S}\right)}$ $⇒$P(Z) = $\frac{1}{8}$

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