Explanation:
Number of atoms that decay in time \(t\)
\(={N}_{0}-{N}_{0} {e}^{-\lambda t}={N}_{0}\left(1-{e}^{-\lambda t}\right)\)
\(\therefore\) Probability that a nucleus decays
\( = \frac{{{\text{ Number }}\,{\text{of}}\,{\text{ nuclei decayed }}}}{{{\text{ total }}\,{\text{number }}\,{\text{of }}\,{\text{nuclei }}}}\)
\( = \frac{{{N_0}\left( {1 - {e^{ - \lambda t}}} \right)}}{{{N_0}}}\)
\(\therefore \quad {P}\) (probability) \(=1-{e}^{-\lambda t}\)
But \(\lambda=\frac{\ln 2}{{~T}}\)
\(\therefore {\mkern 1mu} \,\,\,\,P = 1 - {e^{ - \ln 2\left( {\frac{1}{{\;T}}} \right)}}\)
Here, \(P = 1 - {e^{ - \ln 2(20/365)}}\)
\( = 1 - {e^{\frac{{ - 0.693 \times 4}}{{73}}}}\)
\( = 1 - {e^{ - 0.038}}\)
\( = 1 - 0.963\)
\( = 0.037\)
\( = 3.7\% \)