A) \[\frac{3}{7}\]
B) \[\frac{1}{7}\]
C) \[\frac{3}{8}\]
D) \[\frac{7}{8}\]
Correct Answer: C
Solution :
Total number of cases \[n(S)={{2}^{3}}=8\] Favorable cases \[=\{(H,T,T),(T,H,T),(T,T,H)\}\] \[n(F)=3\] \[\therefore \]Required probability \[=\frac{3}{8}\] Alternate Solution: Probability of getting head in one coin is \[p=\frac{1}{2}\] \[\Rightarrow \,\,\,\,\,q=\frac{1}{2}\] \[\therefore \]Probability of getting one head in three tosses \[={{\,}^{3}}{{C}_{1}}{{\left( \frac{1}{2} \right)}^{1}}{{\left( \frac{1}{2} \right)}^{2}}\] \[=3{{\left( \frac{1}{2} \right)}^{3}}\] \[=\frac{3}{8}\]
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