A) 3
B) 5
C) 2
D) 4
Correct Answer: A
Solution :
Probability of at least one successes in k trials is \[\sum\limits_{r=1}^{K}{^{k}{{C}_{r}}\,{{p}^{r}}{{d}^{k-1}}}\] When \[k=3\] \[=\sum\limits_{r=1}^{3}{^{3}{{C}_{r}}\,{{P}^{r}}{{d}^{3-r}}}\] (where \[p=\frac{2}{5},\,\,q=\frac{3}{5}\]) \[{{=}^{3}}{{C}_{1}}\,{{\left( \frac{2}{5} \right)}^{1}}{{\left( \frac{3}{5} \right)}^{2}}{{+}^{3}}{{C}_{2}}{{\left( \frac{2}{5} \right)}^{2}}\left( \frac{3}{5} \right){{+}^{3}}{{C}_{3}}{{\left( \frac{2}{5} \right)}^{3}}{{\left( \frac{3}{5} \right)}^{0}}\] \[=3\left( \frac{2}{5} \right)\times \frac{9}{25}+3\times \frac{4}{25}\times \frac{3}{5}+\frac{8}{125}\times 1\] \[=\frac{54}{125}+\frac{36}{125}+\frac{8}{125}\] \[\frac{98}{125}>\frac{7}{10}\]
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