Three coins are tossed together. Find the probability of getting:
(i) exactly two heads
(ii) at least two heads
(iii) at least one head and one tail
(iv) no tails
When 3 coins are tossed together, the outcomes are as follows:
$\mathrm{S}=\{(\mathrm{h}, \mathrm{h}, \mathrm{h}),(\mathrm{h}, \mathrm{h}, \mathrm{t}),(\mathrm{h}, \mathrm{t}, \mathrm{h}),(\mathrm{h}, \mathrm{t}, \mathrm{t}),(\mathrm{t}, \mathrm{h}, \mathrm{h}),(\mathrm{t}, \mathrm{h}, \mathrm{t}),(\mathrm{t}, \mathrm{t}, \mathrm{h}),(\mathrm{t}, \mathrm{t}, \mathrm{t})\}$
Therefore, the total number of outcomes is 8 .
(i) Let A be the event of getting triplets having exactly 2 heads.
Triplets having exactly 2 heads: $(\mathrm{h}, \mathrm{h}, \mathrm{t}),(\mathrm{h}, \mathrm{t}, \mathrm{h}),(\mathrm{t}, \mathrm{h}, \mathrm{h})$
Therefore, the total number of favourable outcomes is 3 .
$\mathrm{P}(A)=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}=\frac{3}{8}$
(ii) Let $A$ be the event of getting triplets having at least 2 heads.
Triplets having at least 2 heads: $(\mathrm{h}, \mathrm{h}, \mathrm{t}),(\mathrm{h}, \mathrm{t}, \mathrm{h}),(\mathrm{t}, \mathrm{h}, \mathrm{h}),(\mathrm{h}, \mathrm{h}, \mathrm{h})$
Therefore, the total number of favourable outcomes is 4 .
$\mathrm{P}(A)=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}=\frac{4}{8}=\frac{1}{2}$
(iii) Let A be the event of getting triplets having at least one head and one tail.
Triplets having at least one head and one tail: $(\mathrm{h}, \mathrm{h}, \mathrm{t}),(\mathrm{h}, \mathrm{t}, \mathrm{h}),(\mathrm{t}, \mathrm{h}, \mathrm{h}),(\mathrm{h}, \mathrm{h}, \mathrm{t}),(\mathrm{t}, \mathrm{t}, \mathrm{h}),(\mathrm{t}, \mathrm{h}, \mathrm{t})$
Therefore, the total number of favourable outcomes is 6 .
$\mathrm{P}(A)=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}=\frac{6}{8}=\frac{3}{4}$
(iv) Let $A$ be the event of getting triplets having no tail.
Triplets having no tail: $(\mathrm{h}, \mathrm{h}, \mathrm{h})$
Therefore, the total number of favourable outcomes is $1 .$
$\mathrm{P}(A)=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}=\frac{1}{8}$
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