We have two coins, A and B. For each toss of coin A, the probability of getting head is 1/2 and for each toss of coin B, the probability of getting Heads is 1/3. All tosses of the same coin are independent. We select a coin at random and toss it till we get a head. The probability of selecting coin A is ¼ and coin B is 3/4. What is the expected number of tosses to get the first heads?
The above problem is taken from the website https://www.analyticsvidhya.com/blog/2017/04/40-questions-on-probability-for-all-aspiring-data-scientists/ question 11
My solution is either 1/(1/4*1/2 + 3/4*1/3)=8/3, including the success toss, or 5/3, not including the success toss. My understanding is that it is geometric distribution question.
But the solution provided is 2.75, with the following explanation: "If coin A is selected then the number of times the coin would be tossed for a guaranteed Heads is 2, similarly, for coin B it is 3. Thus the number of times would be Tosses = 2 * (1/4)[probability of selecting coin A] + 3*(3/4)[probability of selecting coin B] = 2.75"
Is the solution provided incorrect? Or am I missing something?