Question on Probabilty Percentages

[taskforce141]

Hello. The following is the problem for the question. I already tried it myself and think I got a good solution. Can anyone just check my results to make sure I have done everything correctly? Thank you very much in advance.

Homework Statement

An industrial plant is conducting a study to determine how quickly injured workers are back on the job following injury. Records show that 10% of all injured workers are admitted to the hospital for treatment and 15% are back on the job the next day. In addition, studies show that 2% are both admitted for hospital treatment and back on the job the next day. If a worker is injured, what is the probability that the worker will either be admitted to a hospital or be back on the job the next day or both?2. The attempt at a solution
P(T) = 0.1 (Admitted to hospital)
P(N) = 0.15 (Back to work the next day)
P(T|N) = 0.2 (Both)
P(S) = ?

P(S) = P(T) + P(N) - P(T|N)
= 0.1 + 0.15 - 0.02 = 0.23

 

[Ray Vickson]

Hello. The following is the problem for the question. I already tried it myself and think I got a good solution. Can anyone just check my results to make sure I have done everything correctly? Thank you very much in advance.

Homework Statement

An industrial plant is conducting a study to determine how quickly injured workers are back on the job following injury. Records show that 10% of all injured workers are admitted to the hospital for treatment and 15% are back on the job the next day. In addition, studies show that 2% are both admitted for hospital treatment and back on the job the next day. If a worker is injured, what is the probability that the worker will either be admitted to a hospital or be back on the job the next day or both?


2. The attempt at a solution
P(T) = 0.1 (Admitted to hospital)
P(N) = 0.15 (Back to work the next day)
P(T|N) = 0.2 (Both)
P(S) = ?

P(S) = P(T) + P(N) - P(T|N)
= 0.1 + 0.15 - 0.02 = 0.23

Your understanding of the concepts seems faulty. P(T|N) denotes the *conditional* probability of T, given that N occurs; in words, that would be the fraction of those workers who return the next day that are admitted to hospital. That is NOT what you were told. You were told the fraction who are both admitted and return the next day, that is, P(N & T). You were also told this is 2%, but you have listed it as 0.2, which is 20%.

In actual fact, P(T|N) = P(T & N)/P(N) = 0.02/0.15 = 2/15, which is nowhere near your figure of 0.2.

In your last line you did compute P(S) properly, but I am concerned about your previous mistakes and wonder whether you really did understand what you were doing in the last computation.

RGV

 

[taskforce141]

Thank you very much for your help. I now realize that I have mistakenly put 0.2, but if you see in the calculation I used the correct value 0.02.

I understand that I didn't use the correct notation in my calculations. Instead of P(T|N), I meant to use P(T & N) so I would then be able to use the Additive Rule,
P(A Or B) = P(A) + P(B) - P(A & B). It seems I accidentally wrote down the wrong thing. All of this aside, I appreciate you checking my work. Thanks.

 

[Ray Vickson]

Thank you very much for your help. I now realize that I have mistakenly put 0.2, but if you see in the calculation I used the correct value 0.02.

I understand that I didn't use the correct notation in my calculations. Instead of P(T|N), I meant to use P(T & N) so I would then be able to use the Additive Rule,
P(A Or B) = P(A) + P(B) - P(A & B). It seems I accidentally wrote down the wrong thing. All of this aside, I appreciate you checking my work. Thanks.

Good. Glad to hear it (although, I suppose since I am reading this on a screen I am not really hearing it---but you know what I mean).

RGV