A) 64
B) 21
C) 32
D) 42
Correct Answer: D
Solution :
[d] Here \[P=\frac{1}{2}.q=1-p=1-\frac{1}{2}=\frac{1}{2}\] n=6, N=64. Then \[P(r){{=}^{n}}{{C}_{r}}{{p}^{r}}{{q}^{n-1}}{{=}^{6}}{{C}_{r}}{{\left( \frac{1}{2} \right)}^{r}}.{{\left( \frac{1}{2} \right)}^{6-r}}{{=}^{6}}{{C}_{r}}{{\left( \frac{1}{2} \right)}^{6}}\] \[\therefore f(r)=Np(r)={{64.}^{6}}{{C}_{r}}.\frac{1}{64}{{=}^{6}}{{C}_{r}}\] Now \[\sum\limits_{3}^{6}{p(r)=p(3)+p(4)+p(5)+p(6)}\] \[={{(}^{6}}{{C}_{3}}{{+}^{6}}{{C}_{4}}{{+}^{6}}{{C}_{6}})\frac{1}{{{2}^{6}}}\] \[=({{2}^{6}}{{-}^{6}}{{C}_{0}}{{-}^{6}}{{C}_{1}}{{-}^{6}}{{C}_{2}})\frac{1}{{{2}^{6}}}\] \[=(64-1-6-15)\frac{1}{{{2}^{6}}}=\frac{42}{64}=\frac{21}{32}\] \[\therefore \,\,\,f(r){{\,}_{r\ge 3}}=N\sum\limits_{3}^{6}{p(r)=64.}\frac{21}{32}=42\]
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