Probability that a game show contetstent picks a red ball from box 1

Question

A game involves picking coloured balls from two boxes, referred to as Box 1 and Box 2. Box 1 contains two red balls and seven green balls. Box 2 contains four green balls and three red balls. The balls are identical in every respect except for their colour. A blindfolded player first chooses a box and then picks a ball. If a certain player picks a red ball, what is the probability that the player chose from Box 1?

The formula I have used to solve this is Bayes Theron

Plugging in the numbers asked in the question

= .2/ or 20%

Is this the correct answer also did I use the formula correctly?

Answer 1

$p(1\text{ and }R)=\frac{1}{2}\cdot\frac{2}{9}=\frac{1}{9}$. $p(2\text{ and }R)=\frac{1}{2}\cdot\frac{3}{7}=\frac{3}{14}$. Hence $$p(1|R)=\frac{\frac{1}{9}}{\frac{1}{9}+\frac{3}{14}}=\frac{14}{41}$$

Answer 2

Let $R$ be the probability he picks a red ball, $I$ for box 1, $II$ for box 2.

So by the law of total probability $$P(R) = P(R|I)P(I) + P(R|II)P(II) = \frac{2}{9}\frac12 + \frac37\frac12 = \frac{7}{63} + \frac{27}{126}= \frac{41}{126}$$

and $$P(R \cap I)=P(R|I)P(I) = \frac12\cdot \frac29=\frac{2}{18} = \frac{14}{126}$$

so that $$P(I | R)=\frac{P(I \cap R)}{P(R)} = \frac{14}{41} $$