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A fair coin is tossed $3$ times. Find the probability of :-

(a) throwing $3$ heads, given that the first toss is a head.

(b) throwing $3$ heads, given that the first two tosses result in heads.

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closed as off-topic by Dragonemperor42, Davide Giraudo, Chris Godsil, C. Falcon, Leucippus Apr 24 '17 at 3:18

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  • $\begingroup$ What have you tried? Hint: If your first toss is a head, then your sample space is of the form $H,_,_$ ; where _ can be either H or T. $\endgroup$ – Dragonemperor42 Apr 23 '17 at 17:21
  • $\begingroup$ @elizabeth-han i tried to use condtional formula but my answer is 1/8 and the answer of book is 1/4 $\endgroup$ – user373141 Apr 23 '17 at 17:24
  • $\begingroup$ @prayersmith I assume you're talking about part A. What's the conditional formula and how did you apply it? $\endgroup$ – Elizabeth Han Apr 23 '17 at 17:28
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(a) Note that if $A$ designates event that you throw 3 heads and $B$ designates event that your first throw is head, then $A\cap B$ =$A$, because $B\subseteq A$

So $P(A\cap B)=P(A)=(1/2)^3$

$P(B)=1/2\cdot1\cdot1$

Now $P(A|B)=\frac{P(A\cap B)}{P(B)}=1/4$

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This is just a classic example of the gambler's fallacy. You know that the flips of the coins that have already occurred will not affect the probability of your next flip. Therefore, you can just ignore the flips that have already happened and simulate the sequence from after those flips. Can you figure it out from here?

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