Suppose you have a fair six-sided die, one fair coin, and one biased coin. The biased coin has probability 0.25 of showing Heads. First you roll the die. If the number on the die is less than three, you then toss the fair coin until you see a Head; otherwise, you toss the biased coin until you see a Head. Compute the expectation of the number of tosses, giving your answer correct to two decimal places.
I have no clue how to approach this.any hints would be helpful
On
If the probability of showing a Head (H) in a flip of a coin is $p$ then expected number of flips to see the first H is $\displaystyle \frac{1}{p}$.
Using that, find expected number of flips $\small E(F)$ for the fair coin and expected number of flips for the biased coin $\small E(B)$.
Now probability of die showing less than $3, \small P(X \lt 3)= \displaystyle \frac{2}{6}$ (die shows either $1$ or $2$)
So expected number of flips to see first H $ \small = P(X \lt 3) \times E(F) + (1-P(X \lt3)) \times E(B)$
Based on your comments, I would suggest revising the concepts of expectation values before attempting this question. Nevertheless, I have provided a solution below with some references and definitions to guide you towards the correct path. I haven't provided the explanation for everything so that you spend some time understanding the concepts and finally figure it out on your own(if you need some more explanation or are stuck somewhere, please reach out via the comment section). Hope it helps.