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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}

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For the first question, you are not asked to find the probability of obtaining four heads in a row (which is indeed $1/16$). You are given that the boy tossed three heads in a row and asked to find the probability that his next toss is a head. Can you calculate this? –  David Mitra Sep 2 '12 at 17:58
    
so "again" doesn't mean anything? I thought due to "again"; question implies that all heads case should be considered. The "next" heads probability is 1/2; what about second question? –  Anubhav Saini Sep 2 '12 at 18:03
    
"Again" would imply that all four tosses were heads of course; but still you are given that the first three tosses were heads. The probability that the fourth toss is a head, given that the first three were, is $1/2$, as you said. Your reasoning for the second question in the next to last paragraph of your post is correct. –  David Mitra Sep 2 '12 at 18:07
1  
Referring to your EDIT 1: Yes, your answers to Q3, Q4, Q5 are all correct. –  Hagen von Eitzen Sep 4 '12 at 5:25

1 Answer 1

up vote 5 down vote accepted

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.

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