- #1
Azherbahai
- 5
- 0
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.