 If you flip a coin 9 times, you get a John Stewart 2022-01-06 Answered
If you flip a coin 9 times, you get a sequence of heads (H) and tails (T).
(a) How many different sequences of heads and tails are possible?
(c) How many different sequences have at most 2 heads?

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Use Product rule: If one event can occur in m ways AND a second event can occur in n ways , then the number of ways that the two events can occur in sequence is then $$\displaystyle{m}\times{n}$$.
Definition of permutation:
$$\displaystyle{P}{\left({n},{r}\right)}={\frac{{{n}!}}{{{\left({n}−{r}\right)}!}}}$$
Definition of combination:
$$\displaystyle{C}{\left({n},{r}\right)}={\left({\frac{{{n}}}{{{r}}}}\right)}={\frac{{{n}!}}{{{r}!{\left({n}−{r}\right)}!}}}$$
With $$\displaystyle{n}\ne{n}\times{\left({n}-{1}\right)}\times\ldots\times{2}\times{1}$$
(a) If the coin is flipped 9 times and each flip has 2 possible outcomes, we can use the product rule:
$$\displaystyle{2}\times{2}\times{2}\times{2}\times{2}\times{2}\times{2}\times{2}\times{2}={29}={512}$$
(b)The order of the heads or tails is not important (we are interested in the number of heads, but not the order of the heads), hence we need to use the definition of combination.
$$\displaystyle{n}={9}$$
$$\displaystyle{r}={5}$$
Evaluate the definition of a combination:
$$\displaystyle{C}{\left({9},{5}\right)}={\frac{{{9}!}}{{{5}!{\left({9}−{5}\right)}!}}}={\frac{{{9}!}}{{{5}!{4}!}}}$$
$$\displaystyle{C}{\left({9},{5}\right)}={126}$$
(c)$$\displaystyle{n}={9}$$
$$\displaystyle{r}\leq{2}$$
Evaluate the definition of a combination:
$$\displaystyle{C}{\left({9},{0}\right)}={\frac{{{9}!}}{{{0}!{\left({9}−{0}\right)}!}}}={1}$$
$$\displaystyle{C}{\left({9},{1}\right)}={\frac{{{9}!}}{{{1}!{\left({9}−{1}\right)}!}}}={9}$$
$$\displaystyle{C}{\left({9},{2}\right)}={\frac{{{9}!}}{{{2}!{\left({9}−{2}\right)}!}}}={36}$$
Add the number of outcomes for each value of r :
$$\displaystyle{1}+{9}+{36}={46}$$