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Q.
When the three coins are tossed simultaneously, then the probability of getting one head will be:
Bihar CECEBihar CECE 2005
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} $