Bag I contains 3 red and 4 black balls while another Bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the bags and it is found to be red. Find the probability that it was drawn from Bag II

Question

anonymous · Answer

This is an application of Bayes' theorem

anonymous · Answer

I think you can solve this in one of two ways (though in essence, I think they're the same way)

What you know is that you have drawn a red ball. Out of the total possible outcomes (3+4+5+6=18), 8 of them involve pulling a red ball. How many of those 8 outcomes come from drawing a ball from bag 2?

The other way is to use the formula:
$$P(2|R)=\frac{P(2\cap R)}{P(R)}$$

anonymous · Answer

"2" representing bag 2, and R representing red balls.

anonymous · Answer

Yes the other way is better known as Bayes' rule

anonymous · Answer

35/68 is the answer?

diyadiya · Answer

Yeah

anonymous · Answer

btw, I made a mistake earlier $ P(R) = \frac12 \times \frac37 + \frac 12 \times \frac 5{11} $ and the final answer as consistent to bayes' theorem is $$ \huge \frac{\frac 12 \times \frac 5{11} } {   \frac12 \times \frac37 + \frac 12 \times \frac 5{11} }=\frac{35}{68} $$