I need a hint on the following question:
There are 7 red counters in a bag. The rest of the counters in the bag are blue. There are more blue counters than red counters in the bag. Two counters are taken at random from the bag. The probability that there will be one counter of each colour is $7/15$.
Work out the total number of counters in the bag before they are picked out.
So far I have:
Let x be the number of counters in the bag.
Pr(red counter) is drawn first = 7/x
Pr(blue counter) is drawn first = (x-7)/x
Pr(counters will be same) = 8/15
So I could have the following scenarios:
- Red then Red = 7/x * 8/15
- Red then Blue = 7/x * 7/15
- Blue then Red = (x-7)/x * 7/15
- Blue then Blue = (x-7)x * 8/15
Not sure if I am following the right line of thought here and how to work back from this to get the number of counters i.e. x.
Note that if there are $x$ blue counters, then:
With replacement: $$P_{\text{req}} = \frac7{7+x}\times \frac{x}{7+x} = \frac7{15}$$ Are the solutions real?
Without replacement: $$P_{\text{req}} = P_{\text{blue}_1}P_{\text{red}_2} + P_{\text{red}_1}P_{\text{blue}_2}$$ $$=\frac{x}{7+x}\times \frac7{7+(x-1)} + \frac7{7+x} \times \frac{x}{x+(7-1)} = \frac7{15}$$