Question

In a hospital ward, there are 16 nurses and 5 doctors. 3 of the nurses and 3 of the doctors are male. If a person is randomly selected from this group, what is the probability that the person is male or a doctor?

Answer 1

There are 6/21 chances that the person is male.
There are 5/21 chances that the person is a doctor.
Curiously, there are 3/21 chances that the person is male and a doctor.
Hope I helped!

Answer 2

`\text{P(male or doctors)}=(8)/(21)`

Explanation:

Given : In a hospital ward, there are 16 nurses and 5 doctors. 3 of the nurses and 3 of the doctors are male. If a person is randomly selected from this group.

To find : What is the probability that the person is male or a doctor?

Solution :

According to question,

The outcomes are not mutually exclusive, i.e, one does not exclude the others.

Total number of doctors = 16+5=21

Number of male = 3+3=6

Number of doctors = 5

Number of both male and doctor = 3

`\text{Probability}=\frac{\text{Favorable outcome}}{\text{Total number of outcome}}`

`\text{P(male)}=(6)/(21)`

`\text{P(doctors)}=(5)/(21)`

`\text{P(male doctors)}=(3)/(21)`

In probability - ("OR" → +)

`\text{P(male or doctors)}=\text{P(male)}+\text{P(doctors)}-\text{P(both)}`

If we do not subtract the people who are male and doctors, they will be counted twice.

`\text{P(male or doctors)}=(6)/(21)+(5)/(21)-(3)/(21)`

`\text{P(male or doctors)}=(6+5-3)/(21)`

`\text{P(male or doctors)}=(8)/(21)`