Answer to Question #282260 in Statistics and Probability for John

Question

ST. There are three identical boxes. The first box contains 26 white balls, the second box contains 15 white and Il black balls,

in the third box we have 26 black balls. From the random box we take one white ball. Calculate the probability that a white ball is removed from the first box. 

Answer 1

Let $A,B,C$ be the events of choosing the first box, second box and third box respectively, and let $R$  be the event of drawing a red ball.

Then

$$P(A)=P(B)=P(C)=\dfrac{1}{3},$$

$$P(R|A)=1, P(R|B)=\dfrac{15}{26}, P(R|C)=0$$

By the Bayes' Theorem

$$P(A|R)$$

$$=\dfrac{P(A)P(R|A)}{P(A)P(R|A)+P(B)P(R|B)+P(C)P(R|C)}$$

$$=\dfrac{\dfrac{1}{3}(1)}{\dfrac{1}{3}(1)+\dfrac{1}{3}(\dfrac{15}{26})+\dfrac{1}{3}(0)}=\dfrac{26}{41}$$