A) \[0.649\]
B) \[0.351\]
C) \[0.267\]
D) \[0.667\]
Correct Answer: A
Solution :
p = P (success) = P (getting number greater than 4) \[=P(5,\,6)\] \[=\frac{2}{6}=\frac{1}{3}\] \[\therefore \] \[q=P(failure)=1-P\] \[=1-\frac{1}{3}=\frac{2}{3}\] By Binomial distribution \[P(X=r){{=}^{n}}{{C}_{r}}{{p}^{r}}{{q}^{n-r}}\] P (probability of atleast 2 successes) \[=1-P(X=0,1)\] \[=1-\{(P(0)+P(1)\}\] \[=1-{{\{}^{6}}{{C}_{0}}{{p}^{0}}{{q}^{6}}{{+}^{6}}{{C}_{1}}{{p}^{1}}{{q}^{5}}\}\] \[=1-\left\{ 1{{\left( \frac{1}{3} \right)}^{0}}{{\left( \frac{2}{3} \right)}^{6}}+6\times \left( \frac{1}{3} \right).{{\left( \frac{2}{3} \right)}^{5}} \right\}\] \[=1-\left[ {{\left( \frac{1}{3} \right)}^{0}}\times \frac{{{2}^{6}}}{{{3}^{6}}}+2\times \frac{{{2}^{5}}}{{{3}^{5}}} \right]\] \[=1-\left[ \frac{64}{729}+\frac{2\times 32}{243} \right]=1-\left[ \frac{192+64}{729} \right]\] \[=1-\frac{256}{729}=1-0.35=0.649\]
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