A) \[\frac{3}{214}\]
B) \[\frac{4}{173}\]
C) \[\frac{9}{256}\]
D) \[\frac{7}{231}\]
Correct Answer: C
Solution :
Given, mean = 4 and variance = 2 \[\Rightarrow \] np = 4 and npq = 2 \[n\times 0.3=1-0.778=0.22\] \[n=\frac{0.222}{0.3}=0.74\] \[t=n{{T}_{1/2}}=0.74\times 4.47\times {{10}^{8}}\] \[t=3.3\times {{10}^{8}}yr\] Then, \[Y\left( n,\alpha \right)\] Mean = np = 4 \[\alpha -\] \[_{Z}{{Y}^{A}}{{+}_{0}}{{n}^{1}}{{\to }_{3}}L{{i}^{7}}{{+}_{2}}H{{e}^{4}}\] \[\Rightarrow \] n = 8 \[\therefore \]\[P(X=r){{=}^{n}}{{C}_{r}}{{p}^{r}}{{q}^{n-r}}\] \[{{=}^{8}}{{C}_{r}}{{\left( \frac{1}{2} \right)}^{8}}\]\[\left[ \because p=q=\frac{1}{2} \right]\] The required probability of atleast 7 successes is \[P(X\ge 7)=P(X=7)+P(X=8)\] \[={{(}^{8}}{{C}_{7}}{{+}^{8}}{{C}_{8}}){{\left( \frac{1}{2} \right)}^{8}}\] \[=\left( \frac{8!}{7!1!}+\frac{8!}{8!0!} \right){{\left( \frac{1}{2} \right)}^{8}}\] \[=(8+1){{\left( \frac{1}{2} \right)}^{8}}=\frac{9}{256}\]
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