Suppose a fair coin is tossed nine times. Replace the resulting sequence of H's and 74 Chapter 2 Probability T's with a binary sequence of 1's and 0's (1 for H, 0 for T). For how many sequences of tosses will the decimal corresponding to the observed set of heads and tails exceed 256?

Question

Mathematics

Noah Galvan

15 January, 20:18

Suppose a fair coin is tossed nine times. Replace the resulting sequence of H's and 74 Chapter 2 Probability T's with a binary sequence of 1's and 0's (1 for H, 0 for T). For how many sequences of tosses will the decimal corresponding to the observed set of heads and tails exceed 256?

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Answers (1)

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Laurel Gallagher · Answer 1

255 sequence

Step-by-step explanation:

Since

Head replaced by 1

Tail replaced by 0

For each toss, there are two possible outcomes: heads 1, or tails 0.

For 9 toss the possible outcome sequence range from

000000000 to 111111111

000000000 in binary = 0 in decimal

111111111 in binary = 511 in decimal

Since we are asked to find decimal corresponding to the observed set of heads and tails that exceed 256?

That is sequence that exceed 256 (100000000)

Which is 257 (100000001) to n

So the outcome sequence will range from 257 to the last possible outcome sequence for 9 tosses which is 511

Therefore between 257 to 511.

Number of outcome sequence is 255