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Chapter 2: Problem 2

Let \(X\) represent the difference between the number of heads and the number of tails obtained when a coin is tossed \(n\) times. What are the possible values of \(X ?\)

Short Answer

Expert verified
For even values of \(n\), the possible values of \(X\) are: \[X \in \{ 0, 2, 4, \dots , n \}\] For odd values of \(n\), the possible values of \(X\) are: \[X \in \{ 1, 3, 5, \dots , n \}\]

Step by step solution

01

Identifying the minimum difference

: In the minimum difference case, the number of heads and the number of tails would be equal. This can happen only when we have an even number of tosses (n is even). In this case, we can have half heads and half tails, which makes the minimum difference 0 (zero). If n is odd, it is impossible to have an equal number of heads and tails, so the minimum difference would be 1.
02

Identifying the maximum difference

: The maximum difference between the number of heads and the number of tails would occur when either all tosses result in heads or all tosses result in tails. In this case, X would be equal to the total number of tosses, n.
03

Listing possible values of X

: Now that we know the bounds of X, we can list all possible values within these bounds. For even values of n, the minimum difference in heads and tails is 0, and the maximum difference is n. For odd values of n, the minimum difference is 1, and the maximum difference is still n. In each case, the difference will always be an even number. So, for an even value of n, the possible values of X are: \[x \in \{ 0, 2, 4, \dots , n \}\] For an odd value of n, the possible values of X are: \[x \in \{ 1, 3, 5, \dots , n \}\]

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Most popular questions from this chapter

Suppose the distribution function of \(X\) is given by $$ F(b)=\left\\{\begin{array}{ll} 0, & b<0 \\ \frac{1}{2}, & 0 \leqslant b<1 \\ 1, & 1 \leqslant b<\infty \end{array}\right. $$ What is the probability mass function of \(X ?\)

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