# Conditional probability

A group of students were surveyed by a principal and 2/3 were found to always hand in assignments on time. When questioned about their assignments 3/5 said they always start their assignments on the day they are issued and, of those who always start their assignments on the day they are issues, 11/20 hand them in on time. A) Draw a tree diagram to represent this information
Surveyed group=1
On time =2/3
Not on time=1/3
Always starts=3/5
Not always=2/5
Always start on time=11/20
Always start not on time=9/20
I wish there was option of uploading the solution. Have drawn the tree diagram as explained above.
(edited 2 weeks ago)
Original post by Nashylyn
Surveyed group=1
On time =2/3
Not on time=1/3
Always starts=3/5
Not always=2/5
Always start on time=11/20
Always start not on time=9/20
I wish there was option of uploading the solution. Have drawn the tree diagram as explained above.

i already got the bit you have written, its the second branch where you have to do it for the students who dont start on the day they receive it.
Original post by DWAI
i already got the bit you have written, its the second branch where you have to do it for the students who dont start on the day they receive it.

Yes
Original post by Nashylyn
Yes

how do you do the second branch, i have the solutions but i dont know how the solutions were achieved
From surveyed group branch to on time and not on time. Under on time, branch into always start and not always starts. Under always start branch into on time and not on time. Hope am clear now.
Original post by Nashylyn
From surveyed group branch to on time and not on time. Under on time, branch into always start and not always starts. Under always start branch into on time and not on time. Hope am clear now.

Did you understand this?
Original post by DWAI
i already got the bit you have written, its the second branch where you have to do it for the students who dont start on the day they receive it.

Its a 2x2 tree and you use the info that 3/5 hand their assignment in on time to fill in the partial branch where they dont start their assignment on the first day. The two leaves you need to combine to give the 3/5 are

start their assignment on first day - hand in on time

dont start their assignment on first day - hand in on time

You know the first one and you know the first part of the second one so you can calculate the missing part. Then that gives you the other missing part for

dont start their assignment on first day - dont hand in on time

as probabities sum to 1
(edited 2 weeks ago)

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