MATH1314-1043,1044 16wks SPR21
1 answer:
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Answer:
1) 1 element
2) 13 elements
3) 22 elements
4) 40 elements
Step-by-step explanation:
1) Only one element will have no tails: the event that all the coins are heads.
2) 13 elements will have exactly one tile. Basically you have one element in each position that you can put a tail in.
3) There are elements that have exactly 2 tails. From those elements we have to remove the only element that starts and ends with a tail and in the middle it has heads only and the elements that starts and ends with a head and in the 11 remaining coins there are exactly 2 tails. For the last case, there are
possibilities, thus, the total amount of elements with one tile in the border and another one in the middle is 78-55-1 = 22
4) We can have:
- A pair at the start/end and another tail in the middle (this includes a triple at the start/end)
- One tail at the start/end and a pair in the middle (with heads next to the tail at the start/end)
For the first possibility there are 2 * 11 = 22 possibilities (first decide if the pair starts or ends and then select the remaining tail)
For the second possibility, we have 2*9 = 18 possibilities (first, select if there is a tail at the end or at the start, then put a head next to it and on the other extreme, for the remaining 10 coins, there are 9 possibilities to select 2 cosecutive ones to be tails).
This gives us a total of 18+22 = 40 possibilities.