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s1 probability help!

In a school there are 148 students in Years 12 and 13 studying Science, Humanities or Arts subjects. Of these students, 89 wear
glasses and the others do not. There are 30 Science students of whom 18 wear glasses. The corresponding figures for the Humanities
students are 68 and 44 respectively.

A student is chosen at random.

Find the probability that this student
a is studying Arts subjects,
b does not wear glasses, given that the student is studying Arts subjects.
Amongst the Science students, 80% are right-handed. Corresponding percentages for Humanities and Arts students are 75% and 70%
respectively.
A student is again chosen at random.
c Find the probability that this student is right-handed.
d Given that this student is right-handed, find the probability that the student is studying Science subjects.

Solution: how to do question part d ??

thank you :smile:
Original post by Lamalam

d Given that this student is right-handed, find the probability that the student is studying Science subjects.



So, you're looking for the fraction of the number of students that are "right-handed and study science" over the total number of students that are "right-handed".
(Not worded very well....)

Can you take it from there?
(edited 10 years ago)
Reply 2
Maybe find out how many right handed people study science, then divide by the number of right handed people and multiply by a 100 to give a percentage.
That's how I would do it aha.


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Reply 3
Original post by ghostwalker
So, you're looking for the fraction of the number of students that are "right-handed and study science" over the total number of students that are "right-handed".
(Not worded very well....)

Can you take it from there?


I wrote =

[30/148 + 0.743 ] / (110/148)
=1.27 which is incorrect according to the textbook . why??? @@ correct answer: 0.218

thank you!
Original post by Lamalam
I wrote =

[30/148 + 0.743 ] / (110/148)
=1.27 which is incorrect according to the textbook . why??? @@ correct answer: 0.218

thank you!


OK. Where did you get those figures from? What do they represent?

You're looking for "Number of righthanded science students" / "number of righthanded students"
Reply 5
I've done mine as whole numbers rather than decimals but hopefully you get the idea. ImageUploadedByStudent Room1398240722.120513.jpgQ


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Reply 6
Original post by Lamalam
I wrote =

[30/148 + 0.743 ] / (110/148)
=1.27 which is incorrect according to the textbook . why??? @@ correct answer: 0.218

thank you!


If you calculate a probability that is bigger than 1 then you definitely need to be checking your logic :smile:
Reply 7
Original post by ghostwalker
OK. Where did you get those figures from? What do they represent?

You're looking for "Number of righthanded science students" / "number of righthanded students"


1398492157734.jpg

I think I misused the probability of right hand student instead of the actual no. of right hand people,
Thanks!

One more question :when there is ten balls , 5 red 3 blue others are white , draw three balls from the bag, what is the probability that two balls are red in colour. Is there one method that {RRX}+ {RXR}+ {XRR}, where x can be blue or white, and I havr to multiply thr probability by 3??

Thanks!

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(edited 10 years ago)
Reply 8
Original post by kendellex
I've done mine as whole numbers rather than decimals but hopefully you get the idea. ImageUploadedByStudent Room1398240722.120513.jpgQ


Posted from TSR Mobile


Thank you!!:smile:
Original post by Lamalam

One more question :when there is ten balls , 5 red 3 blue others are white , draw three balls from the bag, what is the probability that two balls are red in colour. Is there one method that {RRX}+ {RXR}+ {XRR}, where x can be blue or white, and I havr to multiply thr probability by 3??


Yes, you only need to work out {RRX} and multiply by 3, as they'll all have the same probability.
Reply 10
Original post by ghostwalker
Yes, you only need to work out {RRX} and multiply by 3, as they'll all have the same probability.

actually why I must multiply the probability by 3? can you explain the logic behind it?
Original post by Lamalam
actually why I must multiply the probability by 3? can you explain the logic behind it?


You can either work out the three probabilities separately {RRX} + {RXR}+ {XRR}, or work out one of them and multiply by 3, as they're numerically all the same.

Order is important here - RRX is a different selection to RXR, and to XRR.
(edited 10 years ago)
Reply 12
Original post by ghostwalker
You can either work out the three probabilities separately {RRX} + {RXR}+ {XRR}, or work out one of them and multiply by 3, as they're mumerically all the same.

Order is important here - RRX is a different selection to RXR, and to XRR.

ok got it :smile:

Thank u!!:smile:

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