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A coin is ﬂipped eight times wright here each ﬂip comes upeither heads or tails. How many type of possible outcomes

a) are tright here in total?

c) contain at leastern three heads?

d) contain the very same number of heads and tails?  First, you have to decide what counts as a possible outcome.

For instance, are \$HHTTTTTT\$ and also \$TTTTTTHH\$ think about various outcomes (the heads and/or tails appear in different orders, wright here order matters?); alternatively, because they both bring about \$2\$ heads and \$6\$ tails, are they taken into consideration the very same outcome (i.e., order does not matter).?

I will certainly assume that the order of the results from each toss does matter, so that in the instance above, we consider the two outconcerns be different.

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(a): How many kind of such outcomes are feasible, given 8 coin tosses?

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

Using the dominion of the product, we have actually that after \$8\$ tosses, the feasible outcomes/sequences is equal to \$\$underbrace 2 imes 2 imes cdots imes 2_huge 8; exttosses = (2)^8= f 256\$\$

(b) Now of these \$256\$ feasible outcomes, we must recognize how many kind of of them are sequences containing

exactly three heads (thus, specifically 5 tails), no fewer, no even more. So tbelow are eight coins, and we must compute exactly how many type of ways we can choose 3 to be heads: \$\$ inom83 = dfrac8!3!,5! = f 56\$\$

at least 3 heads (therefore, at many 5 tails): below we count sequences through specifically 3 heads, and likewise count sequences with even more than 3 heads. What does not count are those sequences containing (zero heads, specifically one head, or precisely 2 heads). We might compute this as \$\$inom 83 + inom 84 + cdots + inom 88\$\$ but it would be far easier to compute its identical, by subtracting from 256 the variety of sequences through no head, only one head, and just 2 heads: \$\$ 256 - left(inom 80 + inom 81 + inom 82 ight)= 256 - (1 + 8 + 28)= f 219\$\$

the same variety of heads and tails: An outcome "counts" if and also just if it contains specifically 4 heads (and therefore, exactly 4 tails). Here, we need only compute the variety of ways to select exactly four heads, since the the other 4 will certainly necessarily then be tails. \$\$f inom 84 = dfrac8!4!,4!= f 70\$\$