The probability that a man  can hit a  target  is ¾ and the probability that he  will hit at least n-1 time is 189/256, then the number of times , n he  tried is 
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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{Probability} \mathrm{that} \mathrm{target} \mathrm{is} \mathrm{hit} \mathrm{is} \mathrm{given} \mathrm{by} \\ \mathrm{p}=\frac{3}{4} \\ \mathrm{Probability} \mathrm{that} \mathrm{target} \mathrm{is} \mathrm{not} \mathrm{hit} \mathrm{is} \mathrm{given} \mathrm{by} \\ \mathrm{q}=1-\frac{3}{4}=\frac{1}{4} \\ \mathrm{Probability} \mathrm{that} \mathrm{target} \mathrm{is} \mathrm{hit} \mathrm{atleast} \mathrm{n}-1, \mathrm{times} \\ =\mathrm{Probabilty} \mathrm{that} \mathrm{target} \mathrm{is} \mathrm{hit} \mathrm{n}-1 \mathrm{times}+\mathrm{Probability} \mathrm{that} \mathrm{target} \mathrm{is} \mathrm{n} \mathrm{times} \\ \Rightarrow \\ \mathrm{Probabilty} \mathrm{that} \mathrm{target} \mathrm{is} \mathrm{hit} \mathrm{n}-1 \mathrm{times}+\mathrm{Probability} \mathrm{that} \mathrm{target} \mathrm{is} \mathrm{n} \mathrm{times}=\frac{189}{256} \\ {\Rightarrow }^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}-1} {\mathrm{p}}^{\mathrm{n}-1} {\mathrm{q}}^1{+}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}{\mathrm{p}}^{\mathrm{n}}=\frac{189}{256} \\ \Rightarrow \mathrm{n} {\left(\frac{3}{4}\right)}^{\mathrm{n}-1} \left(\frac{1}{4}\right)+1\times {\left(\frac{3}{4}\right)}^{\mathrm{n}}=\frac{189}{256} \left[\mathrm{As}{ }^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}-1}=\mathrm{n} \mathrm{and}{ }^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}=1\right] \\ \Rightarrow {\left(\frac{3}{4}\right)}^{\mathrm{n}-1}\left[\frac{\mathrm{n}}{4}+\frac{3}{4}\right]=\frac{189}{256} \\ \Rightarrow \frac{3^{\mathrm{n}-1}}{4^{\mathrm{n}-1}}\left(\frac{\mathrm{n}+3}{4}\right)=\frac{27\times 7}{4^4} \\ \Rightarrow \frac{3^{\mathrm{n}-1}\left(\mathrm{n}+3\right)}{4^{\mathrm{n}}}=\frac{7\times 3^3}{4^4} \\ \Rightarrow \frac{3^{\mathrm{n}-1}\left(\mathrm{n}+3\right)}{4^{\mathrm{n}}}=\frac{\left(4+3\right)\times 3^{\left(4-1\right)}}{4^4} \\ \mathrm{On} \mathrm{comparing} \mathrm{we} \mathrm{get} \mathrm{n}=4 \left(\mathrm{Answer}\right) \end{gathered}$$

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