Die A has four red and two white faces whereas die B has two red and four white faces. A single coin is flipped once. If it falls head the game starts with the throwing of die A and if it falls tail die B is to be used first. The probability of getting a red face at any throw of any die is

Correct option is A

Given:
​$$P(H)=P(T)=\frac{1}{2}\\P(\text{Red}|A)=\frac{4}{6}=\frac{2}{3}\\P(\text{Red}|B)=\frac{2}{6}=\frac{1}{3}$$​​
Formula Used:
​$$P(\text{Red})=P(H)P(\text{Red}|A)+P(T)P(\text{Red}|B)$$​​
Solution:
​$$P(\text{Red})=\frac{1}{2}\times\frac{2}{3}+\frac{1}{2}\times\frac{1}{3}=\frac{1}{3}+\frac{1}{6}=\frac{1}{2}$$​​
Therefore, the probability of getting a red face is $$\frac{1}{2}.$$​​
The correct answer is (a) $$\frac{1}{2}.$$​​