Three coins are tossed once. Find the probability of getting

Solution:

When three coins are tossed once, the sample space is given by

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

$\therefore$ Accordingly,$n(S)=8$

It is known that the probability of an event A is given by

$\mathrm{P}(\mathrm{A})=\frac{\text { Number of outcomes favourable to } \mathrm{A}}{\text { Total number of possible outcomes }}=\frac{n(\mathrm{~A})}{n(\mathrm{~S})}$

(i) Let B be the event of the occurrence of 3 heads. Accordingly, B = {HHH}

$\therefore \mathrm{P}(\mathrm{B})=\frac{n(\mathrm{~B})}{n(\mathrm{~S})}=\frac{1}{8}$

(ii) Let C be the event of the occurrence of 2 heads. Accordingly, C = {HHT, HTH, THH}

$\therefore \mathrm{P}(\mathrm{C})=\frac{n(\mathrm{C})}{n(\mathrm{~S})}=\frac{3}{8}$

(iii) Let D be the event of the occurrence of at least 2 heads.

Accordingly, D = {HHH, HHT, HTH, THH}

$\therefore \mathrm{P}(\mathrm{D})=\frac{n(\mathrm{D})}{n(\mathrm{~S})}=\frac{4}{8}=\frac{1}{2}$

(iv) Let E be the event of the occurrence of at most 2 heads.

Accordingly, E = {HHT, HTH, THH, HTT, THT, TTH, TTT}

$\therefore \mathrm{P}(\mathrm{E})=\frac{n(\mathrm{E})}{n(\mathrm{~S})}=\frac{7}{8}$

(v) Let F be the event of the occurrence of no head.

Accordingly, F = {TTT}

$\therefore \mathrm{P}(\mathrm{G})=\frac{n(\mathrm{G})}{n(\mathrm{~S})}=\frac{1}{8}$

(vii) Let H be the event of the occurrence of exactly 2 tails.

Accordingly, H = {HTT, THT, TTH}

$\therefore \mathrm{P}(\mathrm{H})=\frac{n(\mathrm{H})}{n(\mathrm{~S})}=\frac{3}{8}$

(viii) Let I be the event of the occurrence of no tail.

Accordingly, I = {HHH}

$\therefore \mathrm{P}(\mathrm{I})=\frac{n(\mathrm{I})}{n(\mathrm{~S})}=\frac{1}{8}$

(ix) Let J be the event of the occurrence of at most 2 tails.

Accordingly, I = {HHH, HHT, HTH, THH, HTT, THT, TTH}

$\therefore \mathrm{P}(\mathrm{J})=\frac{n(\mathrm{~J})}{n(\mathrm{~S})}=\frac{7}{8}$