A) \[11/36\]
B) \[35/36\]
C) \[7/12\]
D) none of these
Correct Answer: A
Solution :
The probability of getting\[4,\,\,p=\frac{1}{6}\] The probability of not getting\[4,\,\,q=\frac{5}{6}\] The probability of getting \[4\] at least one throw \[{{=}^{2}}{{C}_{1}}{{\left( \frac{1}{6} \right)}^{1}}{{\left( \frac{5}{6} \right)}^{1}}{{+}^{2}}{{C}_{2}}{{\left( \frac{1}{6} \right)}^{2}}\] \[=\frac{2.5}{{{6}^{2}}}+\frac{1}{{{6}^{2}}}=\frac{11}{36}\] Alternative Solution: Let \[{{E}_{1}}\] and \[{{E}_{2}}\] be the events of getting \[4\] in first and second throw. \[\therefore \]Required probability \[=P({{E}_{1}}\cap {{\bar{E}}_{2}})+P({{\bar{E}}_{1}}\cap {{E}_{2}})+P({{E}_{1}}\cap {{E}_{2}})\] \[=P({{E}_{1}})P({{\bar{E}}_{2}})+P({{\bar{E}}_{1}})P({{E}_{2}})+P({{E}_{1}})P({{E}_{2}})\] \[=\frac{1}{6}\times \frac{5}{6}+\frac{5}{6}\times \frac{1}{6}+\frac{1}{6}\times \frac{1}{6}\] \[=\frac{10}{36}+\frac{1}{36}\] \[=\frac{11}{36}\]
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