foass77W

2021-06-09

If you flip a coin 9 times, you get a sequence of Heads (H) and tails (T).
Solve each question
(a) How many different sequences of heads and tails are possible?
c) How many different sequences have at most 2 heads?

brawnyN

Expert

a)
Since each flip has two possible outcome, {H,T}.
Then, using product rule, the total number of different sequences of heads and tails can be calculated as,
$S={2}^{n}$
$={2}^{9}$
$=512$
Hence, the number of different sequences of heads and tails is 512.
b) The number of different sequences of heads and tails having exactly five heads can be calculated as
${C}_{r}^{n}=\frac{n!}{\left(n-r\right)!r!}$
Here, the value of r is number of exactly five heads
${C}_{5}^{9}=\frac{9!}{\left(9-5\right)!5!}$
$=\frac{9×8×7×6×5!}{4!5!}$
$=\frac{9×8×7×6×5}{4×3×2×1×5!}$
$=9×2×7$
$=126$ Hence, the number of different sequences of heads and tails have exactly five heads is 126
c)
The number of different sequences having at most 2 heads means the number of heads is not more than or equal to 1.
The value of r can be represented as, $r\le 1$
${C}_{0}^{9}+{C}_{1}^{9}=\frac{9!}{\left(9-0\right)!0!}+\frac{9!}{\left(9-1\right)!0!}+\frac{9!}{\left(9-1\right)!1!}$
$=\frac{9!}{9!}+\frac{9×8!}{8!}$
$=1+9=10$
Hence, the number of different sequences having at most 2 heads is 10

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