Q. The records of a hospital show that 10% of the cases of a certain disease are fatal. If 6 patients are suffering from the disease, then the probability that only three will die, is
A) $8748 \times 10^{- 5}$
B) $1458 \times 10^{- 5}$
C) $1458 \times 10^{- 6}$
D) $41 \times 10^{- 6}$
Explanation
(b)
The probability of suffering of a disease is 10%
$p = \frac{10}{100} = \frac{1}{10}\ and\ q = \frac{9}{10}$
Total number of patients, $n = 6$
$\therefore$ Required probability
$= {\ ^{6}C}_{3}\left( \frac{1}{10} \right)^{3}\left( \frac{9}{10} \right)^{3}$
$= \frac{6.5.4}{3.2.1} \times \frac{1}{1000} \times \frac{9 \times 9 \times 9}{1000}$
$= \frac{2}{10^{5}} \times 729 = 1458 \times 10^{- 5}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.