A) \[4\]
B) \[\frac{17}{4}\]
C) \[\frac{17}{8}\]
D) \[\frac{17}{2}\]
Correct Answer: C
Solution :
| [c] Probability that a machine is faulted |
| \[=\frac{1}{4}=P\] |
| Probability that a machine is not faulted |
| \[=1-\frac{1}{4}=\frac{3}{4}=q\] |
| \[\therefore \] Probability that at most two machine is faulted \[=P(X=0)+P(X=1)+P(X=2)\] |
| \[\therefore \,\,\,\,{{\,}^{5}}{{C}_{0}}{{\left( \frac{1}{4} \right)}^{0}}\cdot {{\left( \frac{3}{4} \right)}^{5}}{{+}^{5}}{{C}_{1}}{{\left( \frac{1}{4} \right)}^{1}}\cdot {{\left( \frac{3}{4} \right)}^{4}}+{{\,}^{5}}{{C}_{2}}{{\left( \frac{1}{4} \right)}^{2}}\cdot {{\left( \frac{3}{4} \right)}^{3}}\] |
| \[={{\left( \frac{3}{4} \right)}^{3}}\cdot k\] |
| \[\Rightarrow \,\,{{\left( \frac{3}{4} \right)}^{2}}+5\cdot \frac{1}{4}\cdot \frac{3}{4}+10\cdot {{\left( \frac{1}{4} \right)}^{2}}=k\] |
| \[\therefore \,\,\,k=\frac{10+15+9}{16}=\frac{34}{16}=\frac{17}{8}\] |
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