Simple probability problem A bag contains $(2m+1)\,$ coins. It is known that
$m$ of these coins have a head on both sides and the
remaining coins are fair. A coin is picked up at
random from the bag and tossed. 
If the probability
that the toss results in a head is $14/19 \,$, then $m$ is
equal to ?
How to go about this? What Sample space to consider?
 A: HINT:
Break it into two parts--first find the probability of getting heads with a "guaranteed" coin, then the probability of getting heads with a fair coin.  You can do one or the other, so you add those two probabilities.

Answer:  (mouse over grey spots to see equations)
I'm not entirely sure about the sample space to consider, but the way I would approach this problem is as follows:   
You can pick any one of the $m$ coins that are double-headed, and be guaranteed a head.  So, the probability of getting heads this way is the number of double-headed coins over the total number of coins:  

 $$\displaystyle\frac{m}{2m+1}$$ 

You can pick any one of the remaining $m+1$ coins, and have $50:50$ chances of getting heads or tails.  So, we first find the probability of picking a fair coin, then multiply it by the chances of getting heads with that coin:

 $$\displaystyle\frac{m+1}{2m+1}\cdot\frac{1}{2}$$

So, the overall probability of getting a head is:

 $$\displaystyle\underbrace{\frac{m}{2m+1}}_{\text{double headed}} + \underbrace{\frac{1}{2}\cdot\frac{m+1}{2m+1}}_{\text{flip a fair head}} = \frac{3m+1}{4m+2}$$

We can tell that $\frac{14}{19}$ won't fit this for, but perhaps try a non-reduced form of the fraction.  For example: $\frac{28}{38},\, \frac{42}{57},\, \frac{56}{76},\ldots$
