Submitted by adamtoner on Sat, 06/03/2017 - 21:52

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 mlgiven thatit 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!