Rahul, thanks for the solution. Below I have my questions related to different interpretation of the probabilities listed, and I am including the official answer and solution which I don't understand at all, as it operates with the method I am unaware
Possible exactly two tails: (TTH), (THT) and (HTT)
Each have a probability = (1/2)*(1/2)*(1/2) = 1/8
Can I say, P(not Tail-One Head)=1/2=P(two heads) above => 1/2 means P(One Head) which is enough for us => then 1/8 is equivalent to P(TTH) & P(THT) & P(HTT)
we cancel this
Probability of getting exactly two tails= 3*(1/8) = 3/8
Also, below P(Head)=0 and we must be consistent with the probability tree here, if we start with P(Head) we finish with P(Head)
Possible exactly three tails: (TTT)
Probability of getting exactly three tails = (1/2)*(1/2)*(1/2) = 1/8
Therefore, Probability of getting at least two tails = (3/8) + (1/8) = 4/8 = 1/2
summing up two probabilities P(TTH) & P(THT) & P(HTT) + P(TTT) we get 1/2 + 0 = 1/2
And below is the official explanation of which I don't understand the red color selected part, please shed some light on this
Explanation: The words "at least" signal that you'll be working with more than one probability, then adding them together later. In this case, those multiple probabilities are the probability that two of the three tosses turn up tails, and the other is the probability is that all three of the tosses turn up tails. The only ways the coin tosses can be arranged to make one of those things possible:
H T T
T H T
T T H
T T T
There are four possible outcomes, in the Â…rest three of which two tosses turn up tails, and in the last of which all three turn up tails. The total number of possibilities is the product of the number of possibilities for each toss: (2)(2)(2) = 8
Probability is, as always, the number of desired outcomes divided by possible outcomes: 4/8 = 1/2
