Help with Elementary Probability

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SUMMARY

The discussion centers on calculating probabilities involving two boxes of transistors, specifically focusing on conditional probability and the concept of mutually exclusive events. The user correctly identifies the probabilities for the first two parts of the problem as 5/7 and 2/7, respectively. However, they struggle with the third part, which requires understanding that the selection is without replacement, leading to the correct answer of 11/14. The user learns that the probability of selecting a second N from the box containing 2 N's and 3 P's is not simply 2/5 due to the change in total transistors after the first selection.

PREREQUISITES
  • Understanding of conditional probability, specifically P(A|B).
  • Knowledge of mutually exclusive events in probability.
  • Familiarity with the concept of probability without replacement.
  • Basic skills in solving probability problems involving multiple events.
NEXT STEPS
  • Study the concept of conditional probability in depth, focusing on P(A|B).
  • Learn about the implications of sampling with and without replacement in probability.
  • Explore examples of mutually exclusive events and their calculations.
  • Practice solving complex probability problems from "Mathematical Methods for the Physical Sciences" by Mary Boas.
USEFUL FOR

This discussion is beneficial for students and learners of probability, particularly those studying mathematical methods in physical sciences, as well as educators seeking to clarify concepts related to conditional probability and sampling techniques.

Azherbahai
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Homework Statement



Some Transistors of two different kinds (say N and P) are stored in two boxes. You know that there are 6 N's in one box and that 2 N's and 3 P's got mixed in the other box, but you do not know which box is which.

You select a box and a transistor from it at random and find that it is an N; i) what is the probability that it came from the box with the 6N's ? ii) The other box ? iii) If another transistor is picked form the same box as the first, what is the probability that it is also an N?

My problem is with part iii). The answers for the other two questions I have the correct answer (5/7 and 2/7 respectively)

Homework Equations


[tex]P(A+B)=P(A)+P(B)[/tex] (for mutually exclusive events)
[tex]P_{A}(B)=\frac{P(AB)}{P(A)}[/tex] (conditional probability)
[tex]P(AB)={P(A)}\cdot{P_{A}(B)}= {P(B)}\cdot{P_{B}(A)}[/tex]

The Attempt at a Solution


iii) Probability = P(two N's from 6N box or two N's from other box)= P(two N's from 6N box)+P(two N's from other box) (this is because we have mutually exclusive events)
[tex]P={\frac{5}{7}}\cdot{\frac{1}{1}}+{\frac{2}{7}} \cdot {\frac{2}{5}}[/tex]
[tex]P= \frac{29}{35}[/tex]

The reasoning is as follows: We have two mutually exclusive events (two N's from first box or two N's from second box) so we can either pick an N from the first box and then pick another from said box which means probability of 5/7 OR we may pick an N from the second box first and THEN another N from same box which gives us probability of event as (2/7)(2/5).

unfortunately, the answer is 11/14 but i don't know why or what I am doing wrong. This question is really annoying me. It is from Mathematical Methods for the Physical Sciences by Mary Boas question 3.20 for chapter 15. This is not a homework assignment, I am trying to relearn probability for Stat Mech.

Also, this is my first post so please be gentle.
 
Physics news on Phys.org
In iii) if you have picked the 2N 3P box and you've already picked one N out of it, the probability of the next one being N isn't 2/5.
 
hmmm ... right you are ... I wonder why i didnt catch that. I did not consider that this was without replacement. Cool Thanks a lot dude.
 

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