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 two tails (viii) no tail (ix) at most two tails

Answer 1

When three coins are tossed once, the sample space is given by 
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} 
∴Accordingly, n(S) = 8 
It is known that the probability of an event A is given by
\(P(A)=\frac{\text{Number of outcomes favourable to A}}{\text{Total number of possible outcomes}} =\frac{n(A)}{n(S)}\)
(i) Let B be the event of the occurrence of 3 heads. Accordingly, B = {HHH}

\(∴P(B) =\frac{n(B)}{n(S)}=\frac{1}{8}\)
(ii) Let C be the event of the occurrence of 2 heads. Accordingly, C = {HHT, HTH, THH} 
\(∴P(C) =\frac{n(C)}{n(S)}=\frac{3}{8}\)

(iii) Let D be the event of the occurrence of at least 2 heads. Accordingly, D = {HHH, HHT, HTH, THH} ∴P(D) =n(D)=n(S)=4/8=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}
\(∴P(E) =\frac{n(E)}{n(S)}=\frac{7}{8}\)

(v) Let F be the event of the occurrence of no head. Accordingly, F = {TTT} 
\(∴P(F) =\frac{n(F)}{n(S)}=\frac{1}{8}\)

(vi) Let G be the event of the occurrence of 3 tails. Accordingly, G = {TTT} 
\(∴P(G) =\frac{n(S)}{n(S)}=\frac{1}{8}\)

(vii) Let H be the event of the occurrence of exactly 2 tails.
Accordingly, H = {HTT, THT, TTH} 
\(∴P(H) =\frac{n(H)}{n(S)}=\frac{3}{8}\)

(viii) Let I be the event of the occurrence of no tail. Accordingly, I = {HHH} 
\(∴P(I) =\frac{n(I)}{n(S)}=\frac{1}{8}\)

(ix) Let J be the event of the occurrence of at most 2 tails. Accordingly, I = {HHH, HHT, HTH, THH, HTT, THT, TTH} 
\(∴P(J) =\frac{n(J)}{n(S)}=\frac{7}{8}.\)