A company has 3 divisions X, Y and Z and the number of
employees in each Division is 10, 30 and 50 respectively.
Two people are selected from each division for promotion
interview. Two people will be promoted from the 6 people
selected for interview. Find the probability of a person being
promoted who is employed in
(a) Division X
(b) Division Y
(c) Division Z
My guess
a) 1/15
b) 1/45
c) 1/75
How to make the guess ?
I only know that there are 10C2 x 30C2 x 50C2 = 23,979,375 ways to choose the
6 persons for the interview.
Then, I am at a loss how to continue to answer the question.
This is my guess:
Probability of being promoted: 2/6 = 1/3
Probability of choosing a person from X to be selected for interview: 2/10 = 1/5
Probability of choosing a person from Y to be selected for interview: 2/30 = 1/15
Probability of choosing a person from Z to be selected for interview: 2/50 = 1/25
(a) Probability that a person is selected from X and promoted: 1/3 x 1/5 = 1/15
(b) Probability that a person is selected from Y and promoted: 1/3 x 1/15 = 1/45
(c) Probability that a person is selected from Z and promoted: 1/3 x 1/25 = 1/75
6C2 = 15 ways are for the choosing of two persons out of the 6 interviewed persons for promotion and the 2 persons promoted can come from X and Y, X and Z, X and X , and other possibilities.
But the question is asking to find the probability of one person being promoted is employed in Division X.
So, I think we cannot use 15/45.
Paiseh....I am not very gd in probability..tried to redo and edited my post.. u got answer to that qn?
Originally posted by lianamaster:Paiseh....I am not very gd in probability..tried to redo and edited my post.. u got answer to that qn?
No, I don't have the answer.
Originally posted by Snoopyies:No, I don't have the answer.
Oic.. never mind...we all try and learn from mistakes...
Originally posted by lianamaster:This is my guess:
Probability of being promoted: 2/6 = 1/3
Probability of choosing a person from X to be selected for interview: 2/10 = 1/5
Probability of choosing a person from Y to be selected for interview: 2/30 = 1/15
Probability of choosing a person from Z to be selected for interview: 2/50 = 1/25
(a) Probability that a person is selected from X and promoted: 1/3 x 1/5 = 1/15
(b) Probability that a person is selected from Y and promoted: 1/3 x 1/15 = 1/45
(c) Probability that a person is selected from Z and promoted: 1/3 x 1/25 = 1/75
It seems that the tree diagram is used in your solution.
X ----- S ------ P
------ NP
------ NS
So,
Probability of choosing a person from X to be selected for interview: 2/10 = 1/5
But
Probability of the selected Division X Employee that will be promoted might not be 2/6 = 1/3 because the two promoted employees can come from X and Y, X and Z and X and X and other possiblities.
Hmm....I feel 2 events are independent of each other.
Originally posted by lianamaster:This is my guess:
Probability of being promoted: 2/6 = 1/3
Probability of choosing a person from X to be selected for interview: 2/10 = 1/5
Probability of choosing a person from Y to be selected for interview: 2/30 = 1/15
Probability of choosing a person from Z to be selected for interview: 2/50 = 1/25
(a) Probability that a person is selected from X and promoted: 1/3 x 1/5 = 1/15
(b) Probability that a person is selected from Y and promoted: 1/3 x 1/15 = 1/45
(c) Probability that a person is selected from Z and promoted: 1/3 x 1/25 = 1/75
The question for (a) is to find the probability of one person being promoted is employed in Division X
I think it might not be the same as (a) Probability that a person is selected from X and promoted.
It seems that the question is asking about given the person is promoted and this promoted person comes from Division X.
Sometimes questions might not be too difficult but its the students that think too much and make it too difficult.
Unless theres someone else who can give a new insight into this question, just try what has been proposed and let us know when you are given the answer to this question.
edited...
kristo and dkcx should be correct
I believe the question is asking about the CHANCES of being promoted of an employee in division X, Y and Z respectively.
So.. the initial solutions should be correct.
out of the 6 people: probability of being promoted = 1/3
probablity of an employee being chosen to represent their division is
Div X : 1/5
Div Y : 1/15
Div Z: 1/25
respectively
hence the chance of
a person being promoted in
Div X: is : 1/3 x 1/5 = 1/15
Div Y : 1/3 x 1/15 = 1/45
Div Z : 1/3 x 1/25 = 1/75
Originally posted by Snoopyies:The question for (a) is to find the probability of one person being promoted is employed in Division X
I think it might not be the same as (a) Probability that a person is selected from X and promoted.
It seems that the question is asking about given the person is promoted and this promoted person comes from Division X.
That's what I think... perhaps there are people who can give a more convincing answer..
NOOOOOOOOOOOOOO i dun understanddddddddddddddddddddd
this sec 4 lvl ? T_T
Confirm A-Maths... haha...
The question is about conditional probability.
So, it cannot be an "O" level E.maths simple probability question.
New Syllabus "O" Add. Maths no longer covers probability.
Conditional Probability is covered in H2 Maths.
---------------------------------------------------------------------------------
Conditional Probability of Getting event A given event B. ie P(A/B)
Event B = A person is promoted
Event A = The person who is promoted comes from Division X.
Probability that a person is promoted
= P(XP, YP, ZP)
= 2/10 x 2/6 + 2/30 x 2/6 + 2/50 x 2/6
= 23/225
Probability of a person being promoted who is employed in division X
= P(A/B)
P(XP)
= ------------------
P(XP, YP, ZP)
2/10 x 2/6
= --------------------------------------------
2/10 x 2/6 + 2/30 x 2/6 + 2/50 x 2/6
= 15/23
Originally posted by Lee012lee:The question is about conditional probability.
So, it cannot be an "O" level E.maths simple probability question.
New Syllabus "O" Add. Maths no longer covers probability.
Conditional Probability is covered in H2 Maths.
---------------------------------------------------------------------------------
Conditional Probability of Getting event A given event B. ie P(A/B)
Event B = A person is promoted
Event A = The person who is promoted comes from Division X.
Probability that a person is promoted
= P(XP, YP, ZP)
= 2/10 x 2/6 + 2/30 x 2/6 + 2/50 x 2/6
= 23/225
Probability of a person being promoted who is employed in division X
= P(A/B)
P(XP)
= ------------------
P(XP, YP, ZP)
2/10 x 2/6
= --------------------------------------------
2/10 x 2/6 + 2/30 x 2/6 + 2/50 x 2/6
= 15/23
Sorry, double post.
Originally posted by Lee012lee:The question is about conditional probability.
So, it cannot be an "O" level E.maths simple probability question.
New Syllabus "O" Add. Maths no longer covers probability.
Conditional Probability is covered in H2 Maths.
---------------------------------------------------------------------------------
Just did a checkup on TS and yeah it should be H2 maths since hes in J1 this year.
The answers we provided earlier was for simple probability though which is in sec 4 E maths since i saw that topic while giving tuition so yiha093 should not be so clueless about it.
cher just teach one lesson only , somemore cher only taught us
if x and y = times fraction
if x or y = add fraction
then those dependent / indepnt . blah blah x.x
then i see from up to down i see h2 maths i so happy.
i scroll down see ur posts i sian-half. haha
Originally posted by yiha093:cher just teach one lesson only , somemore cher only taught us
if x and y = times fraction
if x or y = add fraction
then those dependent / indepnt . blah blah x.x
then i see from up to down i see h2 maths i so happy.
i scroll down see ur posts i sian-half. haha
What we did was using your x and y method so we times the fraction. Did u have problems understanding the question or the answers we post? The former is still fine since its a H2 maths problem but if u have problems understanding how we got 1/15, 1/45, 1/75 for the 3 answers then you really need go revise your probability.
Originally posted by dkcx:What we did was using your x and y method so we times the fraction. Did u have problems understanding the question or the answers we post? The former is still fine since its a H2 maths problem but if u have problems understanding how we got 1/15, 1/45, 1/75 for the 3 answers then you really need go revise your probability.
ya i noe. today do the test paper realised it alr... !
ok tmr i look through the qn in more detail
i just scanned thru then say i dunoe =x sorry i noe wrong attitude here x.x