4. When you toss a fair coin for 8 times and record the result as a sequence of H and T(H for head, T for tail) of the face of the coin that is facing up after each toss. Answerthe following questions: (a) How many different outcomes are possible? Forexample, HHHHHHHH, HTHTTIIT are examples of two outcomesHow many different outcomes have 4 heads and 4 tails?How may different outcomes have more heads than tails (e.g., 6 heads and 2 tails, or 5heads and 3 tails,..)?

Question
Asked Nov 2, 2019
4. When you toss a fair coin for 8 times and record the result as a sequence of H and T
(H for head, T for tail) of the face of the coin that is facing up after each toss. Answer
the following questions: (a) How many different outcomes are possible? For
example, HHHHHHHH, HTHTTIIT are examples of two outcomes
How many different outcomes have 4 heads and 4 tails?
How may different outcomes have more heads than tails (e.g., 6 heads and 2 tails, or 5
heads and 3 tails,..)?
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4. When you toss a fair coin for 8 times and record the result as a sequence of H and T (H for head, T for tail) of the face of the coin that is facing up after each toss. Answer the following questions: (a) How many different outcomes are possible? For example, HHHHHHHH, HTHTTIIT are examples of two outcomes How many different outcomes have 4 heads and 4 tails? How may different outcomes have more heads than tails (e.g., 6 heads and 2 tails, or 5 heads and 3 tails,..)?

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Step 1

For each toss, there are two possible outcomes: heads, or tails.

Using the rule of the product, we have that after 8 tosses, the possible outcomes is computed as follows.

2x2x2x.x 2x2 = 28
8 times
=256
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2x2x2x.x 2x2 = 28 8 times =256

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Step 2

An outcome "counts" if and only if it contains exactly 4 heads (and hence, exactly 4 tails). Obt...

8!
C(8,4)
41(8-4)
8!
4! x 4!
= 70
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8! C(8,4) 41(8-4) 8! 4! x 4! = 70

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