Expected Number of Tosses for a Fair Coin: Calculating E(x) for Heads and Tails

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SUMMARY

The expected number of tosses required to obtain the first head when tossing a fair coin is 2. This conclusion is derived from the probability calculations where the probability of heads (p) is 1/2 and the probability of tails (q) is also 1/2. The expected number of tails before the first head is obtained is 1, as calculated using the formula E(x) = 1/(1-q) = 1/p. The discussion emphasizes the importance of clearly showing steps in probability calculations to ensure understanding.

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Homework Statement



Suppose that a fair coin is tossed repeatedly until a head is obtained for the first time.

(a) What is the expected number of tosses that will be required?
(b) What is the expected number of tails that will be obtained before the first head is
obtained?


Homework Equations



E(x)= from -infinity to +infinity (continuous case)
E(x)= for all x (discrete case)

The Attempt at a Solution



p=head=1/2
q=tail=1-p=1/2

Pr(X>1)=1 for n=2,3,...

Pr(X>n)=qn-1

E(x)=1/(1-q)=1/p=2

-if only 2 tosses are expected to obtain a head then only one tail would be obtained before a head

-does this make sense?
 
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helps if you show you steps to make it clear what you did... i'd start like this
\begin{matrix}<br /> P(1) &amp; = P(H) &amp; = 1/2 &amp; &amp;\to P(X \leq 1) = 1/2 &amp;\to P(X&gt;1) = 1/2 \\ <br /> P(2) &amp; = P(TH) &amp;= (1/2)(1/2) &amp;= 1/4 &amp; \to P(X \leq 2) = 3/4 &amp; \to P(X&gt;2) = 1/4\\<br /> P(3) &amp; = P(TTH) &amp;= (1/4)(1/2) &amp;= 1/8 &amp; \to P(X \leq 2) = 7/8 &amp; \to P(X&gt;1) = 1/8<br /> \end{matrix}

which looks like you're close for P(X>n) but not quite, assuming X is number of tosses for a head - should be able to write out P(X=n) from here and set up a sum for the expected value
E(x) = \sum_{n=0}^{\infty} n P(X=n)
 
not too sure what you meant by
Pr(X>1)=1 for n=2,3,...
 

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