 # 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? (b)How many different sequences of heads and tails have exactly fiveheads? c) How many different sequences have at most 2 heads? foass77W 2021-06-09 Answered
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?
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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