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12. Two unbiased coins are tossed. Calculate the probability of getting

a) Exactly two heads

b) At least two tails

c) At most two tails

S = $\left\{\mathrm{HH},\mathrm{HT},\mathrm{TH},\mathrm{TT}\right\}$

Therefore

*n*(S) = 4

F be the event of getting exactly 2 heads.

F = {HH}

*n*(F) = 1

G be the event of getting atleast 2 tails.

G = {TT}

*n*(G) = 1

Let E be the event of getting at-most 2 tails.

then, E = event of getting 0 tail or 1 tail or 2 tail.

or E = $\left\{\mathrm{TT},\mathrm{HT},\mathrm{TH},\mathrm{HH}\right\}$

Therefore,

*n*(E) = 4

Thus,

P (getting exactly 2 heads) = $\frac{n\left(\mathrm{F}\right)}{n\left(\mathrm{S}\right)}=\frac{1}{4}$

P (getting atleast 2 tails) = $\frac{n\left(\mathrm{G}\right)}{n\left(\mathrm{S}\right)}=\frac{1}{4}$

P (getting at-most 2 tails) = $\frac{n\left(\mathrm{E}\right)}{n\left(\mathrm{S}\right)}=\frac{4}{4}=1$

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