Question

Three coins are tossed once. Find the probability of getting
(i) 3 heads (ii) 2 heads (iii) at least 2 heads (iv) at most 2 heads
(v) no head (vi) 3 tails
(vii) exactly 2 tails (viii) no tail
(ix) at most two tails.

Answer 1

Here three coins are tossed once. The sample space S is
$S=\{HHH,HHT,HTH,THH,TTH,THT,HTT,TTT\}$
$\therefore$ n(S) = 8
$\text{(i) Let A be the event of getting 3 heads.}$
A = {HHH}
$\therefore n\left(A\right)=1$
$\text{Thus P(A)}$$=\frac{n\left(A\right)}{n\left(S\right)}=\frac{1}{8}$
$\text{(ii) Let B be the event of getting 2 heads.}$
B = {HHT, HTH, THH,}
$\Rightarrow n\left(B\right)=3$
$\text{Thus P (B)}$ $=\frac{N\left(B\right)}{n\left(S\right)}=\frac{3}{8}$
$\text{(iii) Let C be the event of getting at least 2 heads.}$
C = {HHH, HHT, HTH, THH}
$\Rightarrow n\left(C\right)=4$
$\text{Thus P(C)}$ $=\frac{n\left(C\right)}{n\left(S\right)}=\frac{4}{8}=\frac{1}{2}$
$\text{(iv) Let D be the event of getting at most 2 heads.}$
D = {HHT, HTH, THH, TTH,THT, HTT, TTT}
$\Rightarrow n\left(D\right)=7$
$\text{Thus}P\left(D\right)=\frac{n\left(D\right)}{n\left(S\right)}=\frac{7}{8}.\left(v\right)\text{Let E be the event of getting no head.}E=TTT\Rightarrow \text{n(E)}=1\text{Thus P(E)}=\frac{n\left(E\right)}{n\left(S\right)}=\frac{1}{8}\text{(vi) Let F be the event of getting 3 tails.}F\left\{TTT\right\}\Rightarrow n\left(F\right)=1\text{Thus P(F)}=\frac{n\left(F\right)}{n\left(S\right)}=\frac{1}{8}\text{(vii) Let G be the event of getting exactly two tails.}G=\{TTH,THT,HTT\}\Rightarrow n\left(G\right)=3\text{Thus P (G) =}\frac{n\left(G\right)}{n\left(S\right)}=\frac{3}{8}\text{(viii) Let K be the event of getting no tail.}\text{K}=\left\{HHH\right\}\Rightarrow n\left(K\right)=1\text{Thus P(K)}=\frac{n\left(K\right)}{n\left(S\right)}=\frac{1}{8}.\text{(xi) Let L be the event of getting at most two tails}L=\{HHH,TTh,THT,HTT,HHT,HTH,THH\}\Rightarrow n\left(L\right)=7\text{Thus P(L)}=\frac{n\left(L\right)}{n\left(S\right)}=\frac{7}{8}$