Given: Three coins are tossed together.
Number of possible outcomes is 23 = 8
(i) Let E be the event of getting exactly two heads
E = {(H, H, T) (H, T, H) (T, H, H)}
n (E) = 3
P (E) =\( \dfrac{ n (E) }{ n (S)}\)
= \( \dfrac{3} {8}\)
(ii) Let E be the event of getting at least two heads
E= {(H, H, T) (H, T, H) (T, H, H) (H, H, H)}
n (E)=4
P (E) = \( \dfrac{ n (E) }{ n (S)}\)
= \( \dfrac{4} {8}\)
= \( \dfrac {1}{2}\)
(iii) Let E be the event of getting at least one head and one tail
E = {(H, T, T) (T, H, T) (T, T, H) (H, H, T) (H, T, H) (T, H, H)}
n (E) = 6
P (E) = \( \dfrac{ n (E) }{ n (S)}\)
= \( \dfrac{6} {8}\)
= \( \dfrac {3}{4}\)
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