Question

in a housing society there are 20% senior citizens, 50% adults excluding senior citizens and 30%
children. The incidence of people who fall sick at any point in time are 30% among senior
citizens, 5% among adults and 10% among children. If a person is chosen at random and is found
to be sick, what is the probability that the person is a senior citizen?​

Answer 1

Answer:

The probability is 0.5217 that the person is a senior citizen.

Step-by-step explanation:

P(S) = 20% = 0.2                     P(sick/S) =30% = 0.3

P(A) = 50% = 0.5                    P(sick/A) = 5% = 0.05

P(C) = 30% = 0.3                    P(sick/C) = 10% = 0.10

P(s/sick) $=\frac{\text{P(s and sick)}}{P(sick)}$

$=\frac{\text{P(sick/S).P(S)}}{\text{P(sick/s).P(S)+P(sick/A).P(A)+P(sick/C).P(C)}}$

$=\frac{0.2\times 0.3}{(0.2\times 0.3)+(0.5\times0.05)+(0.3\times 0.10)}$

$=\frac{0.06}{0.06+0.025+0.03}$

$=\frac{0.06}{0.115}$

=0.5217

The probability is 0.5217 that the person is a senior citizen.

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other probability problem : https://brainly.in/question/7971174