Problem:
$n$coins are laid out in front of you.One of the coins is fair, while the other
$n-1$have probability$0 < \lambda < 1$of showing heads.If all
$n$coins are flipped, find the probability of an even amount of heads.
Let $\mathcal{F}$ represent the outcome of the fair coin and $\mathcal{H}$ the total heads from the $n-1$ unfair coins.
This means the total number of heads is given by $\mathcal{F} + \mathcal{H}$ and we are specifically interested in the probability that $\mathcal{F} + \mathcal{H}$ is even.
Now if we condition on the fair coin (assuming that $\mathcal{F} + \mathcal{H}$ is even):
$$P(\mathcal{F} + \mathcal{H}) = P(\mathcal{F} + \mathcal{H} \mid \mathcal{F} = 0)P(\mathcal{F} = 0) + P(\mathcal{F} + \mathcal{H} \mid \mathcal{F} = 1)P(\mathcal{F} = 1)$$
We know that the probability of the outcomes for the fair coin are:
$$P(\mathcal{F} = 0) = P(\mathcal{F} = 1) = 0.5$$
and we also know the outcomes of $\mathcal{F}$ is either 0 or 1, so we can simplify the above to:
$$P(\mathcal{F} + \mathcal{H}) = P(0 + \mathcal{H}) \cdot 0.5 + P(1 + \mathcal{H}) \cdot 0.5$$
In this case where $\mathcal{F} = 0$ we need $\mathcal{H}$ to be even and for $\mathcal{F} = 1$ we need $\mathcal{H}$ to be odd.
This then leaves us with:
$$P(\mathcal{F} + \mathcal{H}) = 0.5 \cdot P(\mathcal{H} \text{ is even}) + 0.5 \cdot P(\mathcal{H} \text{ is odd})$$
Because of the law of total probability both cases of $P(\mathcal{H} \text{ is even})$ and $P(\mathcal{H} \text{ is odd})$ must sum to 1. This means $P(\mathcal{H} \text{ is odd})$ is equivalent to $1 - P(\mathcal{H} \text{ is even})$.
If we put this back into the equation we have:
$$\begin{align*} P(\mathcal{F} + \mathcal{H} \text{ is even}) &= 0.5 \cdot P(\mathcal{H} \text{ is even}) + 0.5 \cdot (1 - P(\mathcal{H} \text{ is even})) \\ &= 0.5 \cdot P(\mathcal{H} \text{ is even}) + 0.5 \cdot 1 - 0.5 \cdot P(\mathcal{H} \text{ is even}) \\ &= 0.5 \cdot P(\mathcal{H} \text{ is even}) + 0.5 - 0.5 \cdot P(\mathcal{H} \text{ is even}) \\ &= [0.5 \cdot P(\mathcal{H} \text{ is even}) - 0.5 \cdot P(\mathcal{H} \text{ is even})] + 0.5 \\ &= 0 + 0.5 \\ &= 0.5 \end{align*}$$
Which means the probability of an even number of heads is 0.5.