# Thread: Probability that X is appointed as a manager

1. ## Probability that X is appointed as a manager

Question:
The chance of three persons Mr X, Y and Z becoming manager of a company are 4:2:3. The probability that bonus scheme shall be introduced if X, Y and Z become managers are 0.3,0.5,0.8 respectively. Find the probability that the bonus scheme will be introduced. What is the probability that X is appointed as a manager?

P(S) = Probability of bonus being introduced.
P(X) = Probability of Mr X being Manager.
P(Y) = Probability of Mr Y being Manager.
P(Z) = Probability of Mr Z being Manager.
P(S/X) = Probability of scheme being introduced when X becomes manager.
P(S/Y) = Probability of scheme being introduced when Y becomes manager.
P(S/Z) = Probability of scheme being introduced when Z becomes manager.

Probability of bonus scheme being introduced
P(S) = P(X).P(S/X) + P(Y).P(S/Y) + P(Z).P(S/Z) = 4(0.3) + 2(0.5) + 3(0.8) = 4.6

Probability that X is appointed as manager
P(S/X) = $\displaystyle \frac{4(0.3)}{4(0.3)+2(0.5)+3(0.8)}$ = 0.26

Is this Right or Wrong solution......Please let me now ........

2. Originally Posted by Stats
Question:
The chance of three persons Mr X, Y and Z becoming manager of a company are 4:2:3. The probability that bonus scheme shall be introduced if X, Y and Z become managers are 0.3,0.5,0.8 respectively. Find the probability that the bonus scheme will be introduced. What is the probability that X is appointed as a manager?

P(S) = Probability of bonus being introduced.
P(X) = Probability of Mr X being Manager.
P(Y) = Probability of Mr Y being Manager.
P(Z) = Probability of Mr Z being Manager.
P(S/X) = Probability of scheme being introduced when X becomes manager.
P(S/Y) = Probability of scheme being introduced when Y becomes manager.
P(S/Z) = Probability of scheme being introduced when Z becomes manager.

Probability of bonus scheme being introduced
P(S) = P(X).P(S/X) + P(Y).P(S/Y) + P(Z).P(S/Z) = 4(0.3) + 2(0.5) + 3(0.8) = 4.6

Probability that X is appointed as manager
P(S/X) = $\displaystyle \frac{4(0.3)}{4(0.3)+2(0.5)+3(0.8)}$ = 0.26

Is this Right or Wrong solution......Please let me now ........
First off - probability answers will always be between 0 and 1, so your P(S) is wrong.

Are these the only candidates for the position of manager?
Assuming they are then..
Turn the ratio into a fraction/decimal/percentage/whatever you like working with probability in. So 4+2+3=9. So X has 4/9ths of a chance to be manager, Y 2/9, Z 3/9.

So the probability of X becoming manager is 4/9 or 44.44 %

Then your method for finding the probability of the bonus seems to be correct, but you have to change the P(X), P(Y), and P(Z) in a similar fashion to what we've just done here..

3. Originally Posted by Unenlightened
First off - probability answers will always be between 0 and 1, so your P(S) is wrong.

Are these the only candidates for the position of manager?
Assuming they are then..
Turn the ratio into a fraction/decimal/percentage/whatever you like working with probability in. So 4+2+3=9. So X has 4/9ths of a chance to be manager, Y 2/9, Z 3/9.

So the probability of X becoming manager is 4/9 or 44.44 %

Then your method for finding the probability of the bonus seems to be correct, but you have to change the P(X), P(Y), and P(Z) in a similar fashion to what we've just done here..

Please can u give me a proper solution of P(S)

4. Why don't you give it a shot yourself first?

Use the same formula you used in your first attempt. But the proper values of P(X), P(Y), and P(Z).

5. ## Is this right?

Originally Posted by Unenlightened
Why don't you give it a shot yourself first?

Use the same formula you used in your first attempt. But the proper values of P(X), P(Y), and P(Z).
P(S) = $\displaystyle \frac{4}{9}(0.3)$+$\displaystyle \frac{2}{9}(0.5)$+$\displaystyle \frac{3}{9}(0.8)$

=0.46 (Is this right?)

6. Originally Posted by Stats
P(S) = $\displaystyle \frac{4}{9}(0.3)$+$\displaystyle \frac{2}{9}(0.5)$+$\displaystyle \frac{3}{9}(0.8)$

=0.46 (Is this right?)
It's right up until where you wrote down 0.46.
4.6/9 is the answer, not 4.6/10..

Originally Posted by Unenlightened
It's right up until where you wrote down 0.46.
4.6/9 is the answer, not 4.6/10..

And P(S/X) = 1.2/0.46 = 2.6

Is this right ?

8. ## P(s/x) =

P(S/X) = Probability of scheme being introduced when X becomes manager. = P(X) * P(S)
= 4/9 * 0.3
=0.133333

Rember stats, Probability of happening of some thing will be never be greater than one.....

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# the probability of X,Y,and Z becoming manager are 4/9,2/9,1/3respectively. the probability that the bonus scheme will be introduced if X,Y and Z become managerare 3/10,1/2, 4/5 respectively. find the probability that the bonus scheme will be introduced?

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