Eight coins are tossed simultaneously. (a) What is the total number of possible

Question

Eight coins are tossed simultaneously.

(a) What is the total number of possible unique outcomes?

(b) What is the total number of outcomes that have the third coin being heads.

(c) What is the total number of outcomes that have the third, fifth, and seventh coins all being heads.

(d) What is the total number of outcomes that the third, fifth, or seventh coins being heads.

I know the answer for (A) is 2^8=256. And then I think the answer for (b) involves permutation? But I am confused as to how to set up the problem. Please show all work and explain your steps when solving the answers. Please solve all parts. (I need to understand this material thoroughly.)

Explanation / Answer

Answer with explanations:

A. 2^8 outcomes, the 8 coins will result in head or tail each.

=256

B. Total outcomes when 3rd is head = 2^7 ( as 3rd result is fixed to head, so you are left with result of 7 coins)

=128

C. If 3rd, 5 th, and 7th are all heads, then rest 5 coins will be 2^5 combinations

3 coins are fixed in terms of result i.e. they will always come out to be heads. Remaining 5 coins have either head or tail as options.

= 32

D. If 3rd , 5th or 7th are heads = All possible combinations - combo where 3rd,5th and 7th are all tail

= 2^8 -1 ( 2^5) = 2^8 - 2^5. ( 2^8 is total combo . 2^5 is no. of combos where 3rd, 5th and 7th are all tails)

=256-32

= 224

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