Reverse probability problem.

Question

Suppose there are two cards A and B. Both of the cards have a yellow side and a green side. When tossed in the air the probability of the yellow side facing up is %31 for Card A and 35% for Card B. A person randomly chooses one of the cards and tosses it in the air 100 times. The card lands with the yellow side facing up 35 times out of the 100 tosses. What is the probability that it is card B? No additional information is provided.

Answer 1

This is a problem of conditional probability.

I assume, that the person chooses card $A$ $50\%$ of the times and card $B$ the other $50\%$ of the times.

We know that the yellow side showed up $35$ times after $100$ tosses and want to calculate the probability that the chosen card is card $B$. Let $\mathbb{P}$ denote the probability measure, then

\begin{equation*} \begin{split} \mathbb{P}(\text{card B}\ |\ 35\text{ times yellow (random choice)}) &= \frac{\mathbb{P}(\text{card B}\ \cap 35\text{ times yellow (random choice)})}{\mathbb{P}(35\text{ times yellow (random choice of cards)})}\\ & = \frac{\mathbb{P}(35*\text{ yellow (random choice)}\ |\ \text{card B})\cdot\mathbb{P}(\text{card B})}{\mathbb{P}(35*\text{ yellow (random choice)})} \end{split} \end{equation*}

Note that $\mathbb{P}(35\text{ times yellow after random choice of cards})=\frac{1}{2}\cdot\mathbb{P}(35 \text{ times yellow with card A}) + \frac{1}{2}\cdot\mathbb{P}(35 \text{ times yellow with card B})$.

We have (using Binomial Distribution) \begin{align}\tag{1} &\mathbb{P}(35 \text{ times yellow with card A})= {100\choose35}\cdot 0.31^{35}\cdot 0.69^{100-35}\approx0.0578.\\ \tag{2} &\text{Similarly, } \mathbb{P}(35 \text{ times yellow with card B})\approx0.0834. \end{align}

Thus $\mathbb{P}(35\text{ times yellow after random choice of cards})\approx0.0706$.

The final answer is thus $\displaystyle\mathbb{P}(\text{card B}\ |\ 35\text{ times yellow})\approx\frac{0.0834\cdot\frac12}{0.0706}\approx0.5906$.