A) \[\frac{n+1}{{{2}^{m+1}}}\]
B) \[\frac{n+2}{{{2}^{m+1}}}\]
C) \[\frac{m+2}{{{2}^{n+1}}}\]
D) None of these
Correct Answer: B
Solution :
Here \[P(H)=P(T)=\frac{1}{2}\] and \[P(X)=1,\] where \[X\] denotes head or tail. If the sequence of \[m\] consecutive heads starts from the first throw, we have \[(HH\,.....m\,\text{times})\,(XX\,......n\,\text{times)}\]. \[\therefore \]Chance of this event \[=\frac{1}{2}.\frac{1}{2}.\frac{1}{2}\,.........m\,\text{times}\,=\frac{1}{{{2}^{m}}}\] \[\because \,\,\,m+1\] and subsequent throws may be head or tail since we are considering at least \[m\] consecutive heads. If the sequence of \[m\] consecutive heads starts from the second throw, the first must be a tail and we have, the chance of this event \[=\frac{1}{2}.\frac{1}{{{2}^{m}}}=\frac{1}{{{2}^{m+1}}}.\] If the sequence of heads starts from \[{{(r+1)}^{th}}\] throw then the first \[(r-1)\] throws may be head or tail but \[{{r}^{th}}\] throw must be a tail and we have, \[(XX\,........(r-1)\,\text{times})\,T\,(HH\,........\,m\,\text{times})\] \[(XX\,.......\,n-\overline{m-r}\,\text{times})\] The chance of this event also \[\frac{1}{2}\times \frac{1}{{{2}^{m}}}=\frac{1}{{{2}^{m+1}}}\] Since all the above events are mutually exclusive, so the required probability \[=\frac{1}{{{2}^{m}}}+\left( \frac{1}{{{2}^{m+1}}}+\frac{1}{{{2}^{m+1}}}+\,....n\,\text{times} \right)\] \[=\frac{1}{{{2}^{m}}}+\frac{n}{{{2}^{m+1}}}=\frac{n+2}{{{2}^{m+1}}}\]. Note : Students should remember this question as a formula.
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