A) \[8748\times {{10}^{-5}}\]
B) \[1458\times {{10}^{-5}}\]
C) \[1458\times {{10}^{-6}}\]
D) \[41\times {{10}^{-6}}\]
Correct Answer: B
Solution :
The probability of suffering from a diseases is\[10%\]. \[ie,\] \[p=\frac{10}{100}=\frac{1}{10}\] and probability of not suffering from a diseases is \[q=1-\frac{1}{10}=\frac{9}{10}\] Total number of patients,\[n=6\] \[\therefore \]Required probability \[{{=}^{6}}{{C}_{3}}{{\left( \frac{1}{10} \right)}^{3}}{{\left( \frac{9}{10} \right)}^{3}}\] \[=\frac{6\cdot 5\cdot 4}{3\cdot 2\cdot 1}\times \frac{1}{1000}\times \frac{9\times 9\times 9}{1000}\] \[=\frac{20}{{{10}^{6}}}\times 729=\frac{2}{{{10}^{5}}}\times 729\] \[=1458\times {{10}^{-5}}\]
You need to login to perform this action.
You will be redirected in
3 sec
