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In January of 2022, an outbreak of the PROBAB-1550 Virus occurred at the Johnaras Hospital in wards A, B and C. It is known that:

• Ward A has 35 patients, 10 percent of whom have the virus,

• Ward B has 70 patients, 15 percent of whom have the virus,

• Ward C has 50 patients, 20 percent of whom have the virus. (1 point)

(a) What is the probability that a randomly selected student from these three wards has the virus? (1 point)

(b) If a randomly selected student from the hospital has the virus, what is the proba- bility that they are in Ward C?

Answer :

A randomly selected student from the hospital has the virus, the probability that they are in Ward C is 0.2.

The solution to the given problem is explained as follows:

(a) What is the probability that a randomly selected student from these three wards has the virus.

The total number of students in the three wards is:

35 + 70 + 50 = 155 students.

Thus, the probability that a randomly selected student from these three wards has the virus is given by:

P(Probab-1550) = P(A U B U C) = P(A) + P(B) + P(C) - P(A ∩ B) - P(B ∩ C) - P(C ∩ A) + P(A ∩ B ∩ C)

WhereP(A) = probability of selecting a student from ward A and

having the virus = 0.1,

P(B) = probability of selecting a student from ward B and

having the virus = 0.15,

P(C) = probability of selecting a student from ward C and

having the virus = 0.2,

P(A ∩ B) = probability of selecting a student from both wards A and B and having the virus.

P(B ∩ C) = probability of selecting a student from both wards B and C and having the virus.

P(C ∩ A) = probability of selecting a student from both wards C and A and having the virus.

P(A ∩ B ∩ C) = probability of selecting a student from all three wards and having the virus = 0.

From the given information:•

Ward A has 35 patients, 10 percent of whom have the virus,•

Ward B has 70 patients, 15 percent of whom have the virus,•

Ward C has 50 patients, 20 percent of whom have the virus

,Thus,P(A) = 35 × 0.1 / 100 = 0.035,

P(B) = 70 × 0.15 / 100 = 0.105,

P(C) = 50 × 0.2 / 100 = 0.10,

And,P(A ∩ B) = 0.035 × 0.105

= 0.00367,P(B ∩ C)

= 0.105 × 0.1 = 0.0105,

P(C ∩ A) = 0.1 × 0.035 = 0.0035,

P(Probab-1550) = 0.035 + 0.105 + 0.1 - 0.00367 - 0.0105 - 0.0035 + 0

= 0.22333

So, the probability that a randomly selected student from these three wards has the virus is 0.22333.

(b) If a randomly selected student from the hospital has the virus, what is the probability that they are in Ward C?

The probability that a randomly selected student from the hospital has the virus is

P(Probab-1550) = 0.22333.

From Bayes’ theorem,

P(C | Probab-1550) = P(Probab-1550 | C) × P(C) / P(Probab-1550)

where,P(C | Probab-1550) is the probability that a randomly selected student from Ward C has the virus,

P(Probab-1550 | C) is the probability that a student from Ward C has the virus,

P(C) is the probability of selecting a student from Ward C.P(Probab-1550 | C) = 0.2

= probability of selecting a student from Ward C and having the virus,

P(C) = 50 / 155 = probability of selecting a student from Ward C,

Therefore,P(C | Probab-1550) = 0.2 × 0.22333 / 0.22333

= 0.2

Thus, if a randomly selected student from the hospital has the virus, the probability that they are in Ward C is 0.2.

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(a) The probability that a randomly selected student from the three wards has the virus is 24%.

(b) The probability that a randomly selected student from the hospital who has the virus is in Ward C is approximately 24.5%

a) The probability that a randomly selected student from the three wards has the virus is: (10% of 35) + (15% of 70) + (20% of 50) = 3.5 + 10.5 + 10 = 24%.Thus, the probability that a randomly selected student from the three wards has the virus is 24%.


b) If a randomly selected student from the hospital has the virus, the probability that they are in Ward C is given by Bayes' theorem. The formula for Bayes' theorem is:P(A|B) = P(B|A) x P(A) / P(B)where:P(A|B) is the probability of event A occurring given that event B has occurred. In this case, A is the event that the student is in Ward C and B is the event that the student has the virus.P(B|A) is the probability of event B occurring given that event A has occurred. In this case, it is the proportion of patients in Ward C who have the virus, which is 20%.P(A) is the probability of event A occurring. In this case, it is the proportion of all patients in the hospital who are in Ward C, which is 50 / (35 + 70 + 50) = 0.2941.P(B) is the probability of event B occurring. In this case, it is the probability of a randomly selected student having the virus, which is 24%.Thus, the probability that a randomly selected student from the hospital who has the virus is in Ward C is:P(A|B) = 0.2 x 0.2941 / 0.24 ≈ 0.245 or 24.5%.Hence, the probability that a randomly selected student from the hospital who has the virus is in Ward C is approximately 24.5%.

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