Question

IN a hospital, there are 200 beds for patients. OF these, 120 are occupied by males and remaining by females. 20% of the males and 40% of the females are suffering from malaria and rest of them for dengue. If a patient is selected at random, find the probability that he/she is a :A female patientB - MAle patientC - male patient suffering from malariaD - Female patient suffering from dengue

Answer 1

Answer on Question #40524 – Math - Probability

IN a hospital, there are 200 beds for patients. OF these, 120 are occupied by males and remaining by females. 20% of the males and 40% of the females are suffering from malaria and rest of them for dengue. If a patient is selected at random, find the probability that he/she is a: A female patient B - Male patient C - male patient suffering from malaria D - Female patient suffering from dengue

Solution

Total number of beds $N = 200$. Number of beds occupied by males $N_{M} = 120$. So, number of beds occupied by females $N_{F} = N - N_{M} = 200 - 120 = 80$. Number of males suffering from malaria is 20% of total males $N_{MM} = \frac{20}{100} \cdot N_{M} = \frac{20}{100} \cdot 120 = 24$. So, number of males suffering from dengue

$N_{MD} = N_{M} - N_{MM} = 120 - 24 = 96$. Number of females suffering from malaria is $40\%$ of total females $N_{FM} = N_{F} \cdot \frac{40}{100} = 80 \cdot \frac{40}{100} = 32$. So, number of females suffering from dengue

$N_{FD} = N_{F} - N_{FM} = 80 - 32 = 48$. Hence,

A) Probability that the patient is a female

$$P(F) = \frac{N_{F}}{N} = \frac{80}{200} = \frac{2}{5}.$$

B) Probability that the patient is male

$$P(M) = \frac{N_{M}}{N} = \frac{120}{200} = \frac{3}{5}.$$

C) Probability that the patient is a male suffering from malaria

$$P(MM) = \frac{N_{MM}}{N} = \frac{24}{200} = \frac{3}{25}.$$

D) Probability that the patient is a female suffering from dengue = 48 / 200 = 6 / 25

$$P(FD) = \frac{N_{FD}}{N} = \frac{48}{200} = \frac{6}{25}.$$