There are two boxes, Box I and Box II. The selection of a box is determined by rolling a fair six-sided die. If the die shows 1 or 2 , Box I is chosen; otherwise, Box II is chosen. Box I contains 3 red and 2 black balls. Box II contains n red and 4 black balls. A ball is drawn at random from the chosen box and is found to be red. If the probability that the drawn ball came from Box I is 1 4 , then the value of n is :

Step-by-step solution

Let E₁ be the event that Box I is chosen, and E₂ be the event that Box II is chosen. Let R be the event that a red ball is drawn. From the given conditions, the probabilities of choosing the boxes are: P(E₁) = 2 6 = 1 3 P(E₂) = 4 6 = 2 3 The conditional probabilities of drawing a red ball from each box are: P(R|E₁) = 3 3+2 = 3 5 P(R|E₂) = n n+4 By Bayes' theorem, the probability that the ball came from Box I given that it is red is: P(E₁|R) = P(E₁)P(R|E₁) P(E₁)P(R|E₁) + P(E₂)P(R|E₂) Substitute the known values into