Question

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

Answer 1

Given: three coins are tossed together 

Formula: P(E) = \(\frac{favourable \ outcomes}{total \ possible \ outcomes}\)

Total number of possible outcomes are 2³= 8 

(i) let E be the event of getting exactly two heads 

E= {(H, H, T) (H, T, H) (T, H, H)} 

n(E) = 3

P(E) = \(\frac{n(E)}{n(S)}\) 

P(E) = \(\frac{3}{8}\) 

(ii) let E be the event of getting at least two heads 

E= {(H, H, T) (H, T, H) (T, H, H) (H, H, H)} 

n(E) = 4

P(E) = \(\frac{n(E)}{n(S)}\) 

P(E) = \(\frac{4}{8}\) = \(\frac{1}{2}\)

(iii) let E be the event of getting at least one head and one tail 

E= {(H, T, T) (T, H, T) (T, T, H) (H, H, T) (H, T, H) (T, H, H)} 

n(E)=6  

P(E) = \(\frac{n(E)}{n(S)}\) 

P(E) = \(\frac{6}{8}\) = \(\frac{3}{4}\)