Submitted by adamtoner on Sat, 06/03/2017 - 21:52
Forums:
Hi, this may seem like a very basic question, but it has thrown me slightly. The question regards working out the probabilities of two distributions and combining them. The way they have added the probabilities I don't get. The hints and answers booklet states the answer to be: $$\frac{0.6 ( b - 0.5 ) + 0.4 ( a - b )}{0.6b + 0.4a} = \frac{0.4a + 0.2b - 0.3}{0.6b + 0.4a}$$
Whereas I got: $$\frac{0.6 ( b - 0.5)}{b} - \frac{0.4 ( a - b )}{a}$$
There is obviously something I'm missing, but I can't find it. Thanks in advance for any help.
A link to the paper is here.
A link to the Hints and Answers document is here.
Probability
I'll have a proper look tomorrow and get back to you!
Conditional probability
You are being asked to find the probability that the bottle contains more than! 500 ml given that it contains less than 505 ml.
Bayes theorem gives us:
\[
P(A|B) = \frac{P(A \cap B)}{P(B)}
\]
So:
\[
P(\text{more than 500} | \text{less than 505}) \\
= \frac{P(\text{more than 500 AND less than 505})}{P(\text{less than 505})}\\
= \frac{P(500 \lt \text{amount} \lt 505)}{P(\text{less than 505})}\\
=\frac{P(\text{skimmed and } 500 \lt \text{amount} \lt 505)+P(\text{full fat and } 500 \lt \text{amount} \lt 505)}{P(\text{skimmed and less than 505})+P(\text{full fat and less than 505})}\\
=\frac{0.6 \times P(\mu \lt X_1 \lt \mu + \tfrac 1 2 \sigma)+ 0.4 \times P( \mu + \tfrac 1 2 \sigma \lt X_2 \lt \mu +\sigma)}{0.6 \times P(X_1 \lt \mu + \tfrac 1 2 \sigma)+ 0.4 \times P(X_2 \lt \mu +\sigma)}\\
=\frac {0.6(b-0.5)+0.4(a-b)}{0.6b+0.4a}
\]
Hope that's clear, please ask if there are any bits you want clarified.
Ah I see now. What an
Ah I see now. What an oversight, thanks so much for helping to clear that up!