Discrete Probability - Flipping a coin 5 times in a row.

Question

Hey guys, so I think the first thing I'm going to do is write out all of the possibilities (Sample Space). But basically since we're flipping the coin 5 times, and there are only 2 possibilities, there should be 2^5 = 32 outcomes (size of Sample Space)?

Also, how can I write out all the possibilities to ensure that I am not missing any? Since they said "By counting" so I'm assuming they want us to write everything out in the Sample Space, and then count it out.

Answer 1

Hint:

These are the possible outcomes when a coin is flipped $5$ times in a row.

Now you can easily find the solutions for part $a),b)$ and $c)$

$$H,H,H,H,H$$ $$H,H,H,H,T $$ $$H,H,H,T,H $$ $$H,H,H,T,T $$ $$H,H,T,H,H $$ $$H,H,T,H,T $$ $$H,H,T,T,H $$ $$H,H,T,T,T $$ $$H,T,H,H,H $$ $$H,T,H,H,T $$ $$H,T,H,T,H $$ $$H,T,H,T,T $$ $$H,T,T,H,H $$ $$H,T,T,H,T $$ $$H,T,T,T,H $$ $$H,T,T,T,T $$ $$T,H,H,H,H $$ $$T,H,H,H,T $$ $$T,H,H,T,H $$ $$T,H,H,T,T $$ $$T,H,T,H,H $$ $$T,H,T,H,T $$ $$T,H,T,T,H $$ $$T,H,T,T,T $$ $$T,T,H,H,H $$ $$T,T,H,H,T $$ $$T,T,H,T,H $$ $$T,T,H,T,T $$ $$T,T,T,H,H $$ $$T,T,T,H,T $$ $$T,T,T,T,H $$ $$T,T,T,T,T $$

Answer 2

I'm dealing with similiar problems at the moment and yes, my general experience is that when it's written 'by couting' it literally means by counting. So just, write all of the possible combinations occuring in the sample space and calculate the probability of each of the stated events, by counting how many times the event occurs and dividing by the size of the sample space.

Your sample space size is correct :)

Answer 3

There are two possibilities for each of the five tosses of the coin, so there are $2^5 = 32$ possible outcomes in your sample space, as you found.

What is the probability that heads never occurs twice in a row?

Your proposed answer of $13/32$ is correct.

If there are four or five heads in the sequence of five coin tosses, at least two heads must be consecutive.

If there are three heads in the sequence of five coin tosses, the only possibility is that the sequence is HTHTH.

There are $\binom{5}{2} = 10$ sequences of five coin tosses with exactly two heads, of which four have consecutive heads (since the first of these consecutive heads must appear in one of the first four positions). Hence, there are $10 - 4 = 6$ sequences of five coin tosses with exactly two heads in which no two heads are consecutive.

In each of the five sequences of coin tosses in which exactly one head appears, no two heads are consecutive.

In the only sequence of five coin tosses in which no heads appear, no two heads are consecutive.

Hence, the number of sequences of five coin tosses in which no two heads are consecutive is $0 + 0 + 1 + 6 + 5 + 1 = 13$, as you found.

What is the probability that neither heads nor tails occurs twice in a row?

Your proposed answer of $1/16$ is correct since there are only two favorable cases: HTHTH and THTHT, which gives the probability $\frac{2}{32} = \frac{1}{16}$.

What is the probability that both heads and tails occur at least twice in a row?

Your proposed answer of $15/16$ is incorrect.

Since $1 - \frac{1}{16} = \frac{15}{16}$, your answer suggests you mistakenly believed that the negation of the statement that neither heads nor tails occurs twice in a row is that both heads and tails occur at least twice in a row. The negation of the statement that neither heads nor tails occurs twice in a row is that at least two heads or at least two tails are consecutive. For instance, the sequences HHTHT and TTTTH both violate the restriction that neither heads nor tails occur twice in a row without satisfying the stronger requirement that both heads and tails occur at least twice in a row.

If both heads and tails occur at least twice in a row, then there are four possibilities:

• there is a block of three consecutive heads and a block of two consecutive tails
• there is a block of three consecutive tails and a block of two consecutive heads
• there is a block of two consecutive heads and a single head that are separated by a block of two consecutive tails
• there is a block of two consecutive tails and a single tail that are separated by a block of two consecutive heads

A block of three consecutive heads and a block of two consecutive tails can occur in two ways, HHHTT and TTHHH. By symmetry, a block of three consecutive tails and two consecutive heads can occur in two ways. A block of two consecutive heads and a single head that are separated by a block of two consecutive tails can occur in two ways, HHTTH and HTTHH. By symmetry, a block of two consecutive tails and a single tail that are separated by a block of two consecutive heads can occur in two ways. Hence, there are $2 + 2 + 2 + 2 = 8$ favorable cases, giving a probability of $\frac{8}{32} = \frac{1}{4}$.

Answer 4

"By counting the elements in the following events" doesn't necessarily imply that you must write out all the possibilities and check every single one to identify which of them satisfy the given conditions. More generally, "counting" could also mean finding a formula which expresses the numbers of possibilities as a function of the number of tosses. As JMoravitz pointed out in his comment, the first approach is infeasible if the number of tosses were as large as $100$. As it turns out, however, it's not too difficult to derive formulae for the numbers of elements in each of the three types of events as a function of the number of tosses. The answers for the special cases of $5$ tosses and $100$ tosses can then be obtained by substituting those numbers into the formulae.

Let $\ h_n\ $ be the number of sequences of $\ n\ $ tosses that have no consecutive heads and end with a head, and $\ t_n\ $ the number of sequences of $\ n\ $ tosses that have no consecutive heads and end with a tail. Then \begin{align} h_n&=t_{n-1}\ ,\\ t_n&=t_{n-1}+h_{n-1}\ ,\\ &=t_{n-1}+t_{n-2}\ ,\\ t_1&=1\ \text{, and}\\ t_2&=2\ .\\ \end{align} It follows that $\ t_n=F_{n+1}\ $, the $\ (n+1)^\text{th}\ $ Fibonacci number. So the number of sequences of $\ n\ $ tosses with no consecutive heads is $\ t_n+h_n=F_{n+2}\ $. For $\ n=5\ $ this gives $\ F_7=13\ $ as the number of such sequences, and for $\ n=100\ $ Wolfram alpha gives $$ F_{102}=927\,372\,692\,193\,078\,999\,176 $$ as the number of such sequences.

The number of sequences of tosses with no consecutive heads or tails is always $2$, because heads and tails must alternate, so the whole sequence is completely determined by the result of the first toss.

The easiest way of getting the number of sequences of $\ n\ $ tosses in which both heads and tail occur at least twice in a row seems to be to use the principle of inclusion-exclusion to obtain the number of elements in its complement. The complement is the union of the set $\ A\ $ of sequences containing no successive heads with the set $\ B\ $ of sequences containing no successive tails. The intersection $\ A\cap B\ $ of these sets is the set of sequences containing no successive heads or tails. Therefore, by the principle of inclusion-exclusion, \begin{align} |A\cup B|&=|A|+|B|-|A\cap B|\\ &=F_{n+2}+F_{n+2}-2\ , \end{align} and the number of sequences of $\ n\ $ tosses in which both heads and tail occur at least twice in a row is therefore \begin{align} &2^n-2F_{n+2}+2\\ =&\,8\hspace{2em}\text{for }n=5\ . \end{align} For $\ n=100\ $ Wolfram alpha gives the number of such sequences as \begin{align} 2^{100}-2F_{102}&+2=\\ &5\,070\,602\,399\,058\,172\,221\,600\,654\,823\,154\ . \end{align}