Answer:
[tex]\text{P(male or doctors)}=\frac{8}{21}[/tex]
Step-by-step 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
[tex]\text{Probability}=\frac{\text{Favorable outcome}}{\text{Total number of outcome}}[/tex]
[tex]\text{P(male)}=\frac{6}{21}[/tex]
[tex]\text{P(doctors)}=\frac{5}{21}[/tex]
[tex]\text{P(male doctors)}=\frac{3}{21}[/tex]
In probability - ("OR" → +)
[tex]\text{P(male or doctors)}=\text{P(male)}+\text{P(doctors)}-\text{P(both)}[/tex]
If we do not subtract the people who are male and doctors, they will be counted twice.
[tex]\text{P(male or doctors)}=\frac{6}{21}+\frac{5}{21}-\frac{3}{21}[/tex]
[tex]\text{P(male or doctors)}=\frac{6+5-3}{21}[/tex]
[tex]\text{P(male or doctors)}=\frac{8}{21}[/tex]
