I saw this question and calculated 1/16 as the answer.
Q1) A lucky boy has been getting heads every time in three tosses of a coin, what is the probability that he will get heads again in the fourth tossing of the coin?
Then, this question came along and I calculated 1/36 as the answer.
Q2) A fair die is tossed twice. Given that a 2 appears on the first toss, the
probability that the sum of the two tosses is 7 equals?
My analogy for first solution was that getting an H is independent event, thus getting four H in a row should have probability equal to (1/2) ^ 4.
For, second question, I thought {2,5} can occur once out of 36 possible cases of the sample space of die rolled twice. But then I thought, since 2 is already been fetched, all I need is a 5 out of {1,2,3,4,5,6}. So answer to second should be 1/6.
Now, I am confused about the terms, I studied mathematics 6 years ago, since I have forgotten terms and their subtleties. It would be really nice if you can provide solutions with explanations.
Edit 1: Just to clarify that I have understood the concept, are these following Questions and Answers right?
Q3) A lucky boy has tossed a coin 3 times, what is the probability that he will get heads again in the fourth tossing of the coin?
A:) 1/2 with sample space: {H,T}
Q4) A lucky boy has tossed a coin 4 times, what is the probability that he will get heads in the fourth tossing of the coin?
A:) 1/2 with event: {HHHH, HHTH, HTHH, THHH, HTTH, THTH, TTHH, TTTH} and sample space { event + 8 more with ending Tail}
Q5) A lucky boy has tossed a coin 4 times, what is the probability that he will get all heads?
A:) 1/16 where event: {HHHH} and sample space {16 possibilities}
Best Answer
The first is a basic concept of probability: We assume such coin experiments to be independent. That means that no matter what happened before, the next throw will produce heads or tails with probability $1\over 2$ each (provided the coin is fair). Therefore, it does not matter if the lucky boy has gotten heads in previous tosses - the next toss is "fresh" and the answer is $1\over 2$ (if we are allowed to assum that the coin is fair).Your answer was to the difeferent question: What is the probability that the next four tosses produce four heads?
Similarly with the second question. You answered the question: What is the probability that the first toss is a 2 and the second toss is a 5? However, that the first toss is a 2 is already given. Hence the actual question is: What is the probability that the second toss is a 5? Answer is $1\over 6$.
Think of it this way: Given that you have tossed three heads in a row, what is the probability that after the next toss you have a total of four tails? Without the info about the first three tosses, one would calculate $1\over 16$. But in reality, four tails is not even possible if you already have three heads.