A fete is being held in a large hall which has 9 doors.
By mistake, the caretaker has left 3 of them locked.
If someone tries a door at random, what is the prob-
ability that it will be:
(i) locked
(ii) not locked
give step wise correct answer
answer is(i) 3/9=1/3
(ii) 1–1/3=2/3
Answers
Answer:
Probability for locked= no of doors locked/ total no of doors
3/9= 1/3
Probability for unlocked= no of doors unlocked/ total no of doors
= (9-3)/9
= 6/9= 2/3
Another way for probability of unlocked doors= 1 - locked probability
= 1 - 1/3= 2/3
Step-by-step explanation:
Given,
Total doors = 9
Locked doors = 3
To Find,
The probability that the door will be locked =?
The probability that the door will not be locked=?
Solution,
Total outcomes = Total doors
Total outcomes = 9
Number of locked doors = 3
Number of open doors = Total doors - Number of locked doors
Number of open doors = 9 - 3
Number of open doors = 6
We know that probability is calculated by dividing favorable outcomes by total outcomes.
The probability that the door will be locked = Number of locked doors / Total doors
The probability that the door will be locked = 3 / 9 = 1 / 3
Similarly, The probability that the door will not be locked = The number of open doors / Total doors
The probability that the door will not be locked = 6 / 9 = 2 / 3
Hence, if someone tries a door at random, the probabilities that it will be locked and not locked will be 1 / 3 and 2 / 3 respectively.