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Question

Suppose that we flip a fair coin 4 times.

(a) What is the probability of at least two consecutive heads are flipped given that the first flip is a heads?

(b) Are the events in the above part independent?

My attempt

(a) H will denote heads, and T denote tails. Since our first flip is already a H, there are $2^3=8$ different ways that the remaining flips can go. Of these, there are four ways we can get two or more consecutive heads: $HHHH$, $HTHH$, $HHTT$, and $HHTH$. I'm not sure if there is a more systematic way to approach this other than listing them out.

Since there are 4 ways that we can get two or more consecutive heads out of the 8 total, the probability is $\frac{4}{8}=\frac{1}{2}$.

(b) The two events are (A) a flip turns out heads (B) that there are because the coin is fair, and the probability of (B) I found by again listing out the ways we can get two or more consecutive heads $HHHH$, $HHHT$, $HHTT$, etc of a total $2^4=16$ possibilities two coins could take. There are 8 ways to get two consecutive heads. Again, I'm not sure if there's a more systematic approach here.

Two events are independent if their product is equal to the intersection of the events. $P(A)*P(B)= \frac{1}{2}*\frac{1}{2} = \frac{1}{4}$, which is not equal to the probability we found in part (a), $P(A\cap B) = \frac{1}{2}$. So the events are not independent.

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  • $\begingroup$ You seem to have missed $HHHT$ from your calculation in (a). That affects the answer to (b), but if it had not then your argument for (b) would have been flawed as $P(B \mid A) = P(B)$ is equivalent to $A$ and $B$ being independent $\endgroup$ – Henry Oct 12 '16 at 23:51
  • $\begingroup$ For part a, the answer is clearly greater than $\frac 12$. There is a $\frac 12$ chance that the second toss is $H$, which immediately gives you two consecutive heads. But even if it comes up $T$ you might get $HTHH$. You left off $HHHT$. $\endgroup$ – lulu Oct 12 '16 at 23:53
  • $\begingroup$ Given that the first throw is heads, you have an insta 50% chance of double heads. If the next one is tails, what's the chance of getting consecutive heads? $\endgroup$ – Cehhiro Oct 13 '16 at 0:56
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(a) What is the probability of at least two consecutive heads are flipped given that the first flip is a heads?

Either you get another head in the second flip, or you get a tail and then two consecutive heads in the remaining flips. $\tfrac 12+\tfrac 18= \frac 58$

Which, are the outcomes: $\rm \mathbf HHTT, \mathbf HHTH, \bbox[wheat, 2pt]{\mathbf HHHT}, \mathbf HHHH$ and $\rm \mathbf HTHH$

(b) Are the events in the above part independent?

The probability of getting heads in the first toss is $1/2$

The probability of getting two consecutive heads is $\binom 42 \tfrac 1{16}$

Even without calculating we can immediately see that the product will not be $5/8$

Intuitively: Because the head in the first flip can be part of two consecutive heads, we should anticipate a dependency; and indeed there are more outcomes possible for "having two consecutive heads" when the first toss is heads than when it is tails.   (vis: Compare the above to : $\rm \mathbf THHT, \mathbf THHH$ and $\rm \mathbf TTHH$ .)

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