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Question: You flip a fair coin four times; these four flips are independent. De fine the events:

A = "the first two flips result (in this order) in HT",

B = "the second and third flips result in TT".

Are these independent or not? (Answer: Not Independent)

Attempt: To prove independence, I need to prove that P(A intersection B) = P(A)*P(B)

P(A):

Sample space is determined by 2^4=16

{HHHH HTHH THHH HTHT

HHHT HTTH TTHH THTH

HHTT HHTH TTTH THHT

HTTT TTTT TTHT THTT}

In this there are 4 events that start with HT,

Pr(A) = 4/16 = 1/4

Pr(B) = 4/16 = 1/4

Pr(A intersection B) = 2/16 = 1/8

Pr(A intersection B) = Pr(A)*Pr(B) 1/8 = 1/4 * 1/4 1/8 != 1/16

So they are not independent.

Am I correct with this approach? I had to google the 4 flip outcome set, is there another method to determine the individual probabilities and intersection probability without having to write out all the possible outcomes?

Question: You flip a fair coin five times; these five flips are independent. De fine the events:

A = "the first three flips result in HHH",

B = "the number of T in these five flips is at least two"

Are these independent or not? (Answer: Not Independent)

Attempt: Again for this, to prove independence, I need to prove that P(A INT B) = P(A)*P(B)

This time its 5 flips, so do I have to write out all 32 possible outcomes for this? Is there any other method I can use to solve the probabilities of Pr(A), Pr(B), and Pr(A intersection B) for this?

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  • $\begingroup$ There is no need to write out all the cases, you can do the count without listing. For your second question, for instance, the only event in $A\cap B$ is $HHHTT$. That makes it easy to check everything. $\endgroup$
    – lulu
    Commented Nov 26, 2018 at 17:08
  • $\begingroup$ As an intuitive approach, imagine that flipping at least 2 T's counts as a win for you. Now knowing that the first three flips are HHH, doesn't that clearly make it less likely for you to win than you were at the beginning of the game? If so, the events are not independent. $\endgroup$
    – Ned
    Commented Nov 26, 2018 at 22:35

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There is no need to write everything out. As you are clearly aware, that becomes impossible (or at least unreasonable) as the numbers involved get bigger.

For your first question, say: The probability that the first two tosses are $HT$ is $\frac 14$ (as you need the first toss to be $H$ and the second to be $T$, and tosses are independent.). The probability that the second and third are $TT$ is also $\frac 14$. But the probability that the first three are $HTT$ is $\frac 18$ by the same reasoning.

Note: the exact same calculation would work if you had a million tosses instead of four. No toss after the third has any bearing on the problem.

Similarly, for your second question: the probability that the string starts $HHH$ is $\frac 18$. The probability that the string has at least two $T's$ is a bit harder, but we can get it from the binomial distribution. We see $$P(B)=\frac 1{2^5}\times \left( \binom 52+\binom 53+\binom 54+\binom 55\right)=\frac {26}{32}$$ (you could also see this by subtracting from the total the strings with no $T$ or exactly one). Of course $$P(A\cap B)=P(HHHTT)=\frac 1{32}\neq \frac 18\times \frac {26}{32}=P(A)\times P(B)$$ so the events are, again, dependent.

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  • $\begingroup$ What is the general formula for the binomial distribution that you used in this case? And can it be applied to most of these coin flip probability questions? $\endgroup$
    – Toby
    Commented Nov 26, 2018 at 17:29
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    $\begingroup$ In general, if you have a binomial process with success probability $p$ and failure probability $q=1-p$, the probability that you will get exactly $k$ successes in $n$ trials is $\binom nk\times p^k\times q^{n-k}$. $\endgroup$
    – lulu
    Commented Nov 26, 2018 at 17:31
  • $\begingroup$ How did you use this formula when you found P(B)? It seems different than the generic one you just posted or am I missing something? $\endgroup$
    – Toby
    Commented Nov 26, 2018 at 17:37
  • $\begingroup$ In order to have at least two $T's$ out of five you must have exactly two, three, four, or five. $\endgroup$
    – lulu
    Commented Nov 26, 2018 at 17:56

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