A) \[1458\times {{10}^{-5}}\]
B) \[1458\times {{10}^{-6}}\]
C) \[41\times {{10}^{-6}}\]
D) \[8748\times {{10}^{-5}}\]
Correct Answer: A
Solution :
Probability of disease is fatal = \[p=10%\] \[p=\frac{10}{100}=\frac{1}{10},\,\,q=\frac{9}{10}\] Number of patients = 6, Number of die cases = 3 \[\therefore \]Probability that 3 will die \[={}^{6}{{C}_{3}}{{\left( \frac{1}{10} \right)}^{3}}{{\left( \frac{9}{10} \right)}^{3}}=1458\times {{10}^{-5}}\].
You need to login to perform this action.
You will be redirected in
3 sec
