A) \[\frac{1}{2}\]
B) \[\frac{1}{3}\]
C) \[\frac{1}{4}\]
D) None of these
Correct Answer: A
Solution :
Let \[X\] be the number of times \[1,\,\,3\] or 4 occur on the die. Then \[X\]follows a binomial distribution with parameter and \[p=\frac{3}{6}=\frac{1}{2}.\] We have \[P(1,\,\,3\] or 4 occur at most \[n\] times on the die) \[=P(0\le X\le n)=P(X=0)+P(X=1)+.....+P(X=n)\] \[={}^{2n+1}{{C}_{0}}{{\left( \frac{1}{2} \right)}^{2n+1}}+{}^{2n+1}{{C}_{1}}{{\left( \frac{1}{2} \right)}^{2n+1}}+.....+{}^{2n+1}{{C}_{n}}{{\left( \frac{1}{2} \right)}^{2n+1}}\] \[=\left[ {}^{2n+1}{{C}_{0}}+{}^{2n+1}{{C}_{1}}+.....+{}^{2n+1}{{C}_{n}} \right]\text{ }{{\left( \frac{1}{2} \right)}^{2n+1}}\] Let \[S={}^{2n+1}{{C}_{0}}+{}^{2n+1}{{C}_{1}}+......+{}^{2n+1}{{C}_{n}}\] \[\Rightarrow 2S=2.{}^{2n+1}{{C}_{0}}+2.{}^{2n+1}{{C}_{1}}+.......+2{}^{2n+1}{{C}_{n}}\] \[=\left( {}^{2n+1}{{C}_{0}}+{}^{2n+1}{{C}_{2n+1}} \right)+\left( {}^{2n+1}{{C}_{1}}+{}^{2n+1}{{C}_{2n}} \right)+.........\] \[+\left( {}^{2n+1}{{C}_{n}}+{}^{2n+1}{{C}_{n+1}} \right)\] \[\Rightarrow S={{2}^{2n}}.\] Hence required probability \[={{2}^{2n}}{{\left( \frac{1}{2} \right)}^{2n+1}}=\frac{1}{2}.\]
You need to login to perform this action.
You will be redirected in
3 sec
