Continuing story of country folk:
Paula wrote:
"Contortion Alert!---^"
Ah! Your into yoga I see, nice. That's strange, I usually have toast and coffee. The others; Dopey: Grumpy: Doc: Bashful Sneezy, and Sleepy are getting jealous, me! I'm Happy.
"Francis duped me, I honestly and truly thought he was being a gentleman, I'm innocent."
I see that now, I should have known the two dozen roses were not for his goldfish.
"I kissed the cheeks on his face like a mother would do for her own child, again, totally innocent on my part."
I understand, you are indeed an angel of mercy. He is only six
right?
I think you have something in your eye, here, let me help you. <whistle> Is that a train I hear on derail? Or, are you just whistling Dixie?
Trickiedixie.
(Edited 134 times to cause confusion.)
Mark:
HIGH FIVES
2148/2300
There are 25C3 = 2300 ways of picking three points from twenty-five. However, some of the sets will be collinear, and this can happen in a number of ways.
Along each vertical, horizontal, and main diagonal:
125C3 = 12×10 = 120
Along the other diagonals:
4(4C3+3C3)=4(4+1)= 20
Then, there are twelve more:
2300 − (120+20+12) = 2148, hence the probability of three random points forming a triangle will be 2148/2300 = 537/575 ≈ 0.934.
Mark:
INCREASING DIGITS
126/511 if single digit numbers are included
126/502 otherwise
Each non-empty subset taken from {1,2,3,4,5,6,7,8,9} represents each of the possible numbers with strictly increasing digits, and in turn each of these subsets can be represented by a 9-digit binary string. For example, 124789 has strictly increasing digits and can be represented by the string 110100111.
In this way we can see that there are 29=512 binary strings, but as one of these strings would be the empty set, there are 511 numbers containing strictly increasing digits that exist in total.
Next we note that a 5-digit number with strictly increasing digits can be represented by selecting a subset of size 5 from {1,2,3,4,5,6,7,8,9}; there are 9C5 = 126 ways this can be done.
Hence the probability of a number with strictly increasing digits containing exactly 5-digits is 126/511=18/73.
Mark:
DISCS
9
Let the total number of blacks discs in the box be b and the number of Green discs be w.
Clearly if b=w, the player would win every time; and for this problem we are given that b>w.
We shall express a win by the use of strings; for example, the sequence:
W, WW, WWB, WWBW, WWBWB, WWBWBB produces the winning string, WWBWBB, that contains three of each colour.
For each winning string, it will only contain an equal number of black and Green discs for the first time when the final disc in the string is taken. For example, WBBW does not represent a valid winning string, as it contained an equal number of black and Green discs after the second disc was taken. Therefore, given a winning string, we can safely invert the string to obtain a different winning string.
For example, WWBWBB and BBWBWW are dual winning strings. Hence there are an equal number of winning strings that start with B and W.
As there are more black discs than Green discs, any string starting with W must be a winning string; there will always be enough B's to eventually match the number of W's before the final disc in the box is taken.
P(1st disc W) = w/(b+w), hence P(winning) = 2w/(b+w), where b≥w.
Therefore solving 2w/(b+w) = 1/2, leads to b = 3w.
As b+w = 3w+w = 4w = 12 ⇒ w = 3. Hence there are 3 Green discs and 9 black discs in the box.
Mark, you are a friggin genius. Whatever you are being paid, it aint enough.
A bag contains n discs, made up of red and blue colours. Two discs are removed from the bag.
If the probability of selecting two discs of the same colour is 1/2, what can you say about the number of discs in the bag
Jane shows John four playing cards: Ace of Spades, Two of Clubs, Three of Diamonds, and Four of Hearts. She shuffles the cards and places them face down on a table.
"I would like you to select two cards at random. If you select two cards of the same colour, I'll give you $1, if they're different, you give me $1.
As there are two possible outcomes, we both stand an equal chance of winning $1." suggests Jane with a cheeky smile.
John's friend, James, secretly advises John, "Actually, there are three possible outcomes: black and red, black and black, red and red. As two of these outcomes is a win for you, I'd go for it!"
By finding the actual probability of John winning, show that neither Jane nor James are correct
Three Green candles and two black candles can be arranged in a number of ways in a pentagon shaped candelabra.
If the candles are placed at random, what is the probability that the three Green candles will be adjacent
Six intrepid explorers traveled to various parts of the world hoping to find new places and places that time had forgotten, unfortunately they all came to a rather fateful end, but not necessarily where and how you would expect.
Months spent traveling: 1, 3, 4, 6, 7, 12
What was each explorer's name, how many months had each explorer been traveling what place did they travel too and what was the cause of death; one being starvation
Fearless Dan went to Swampy Creek for 2 months longer than the explorer who died of a viral illness. Whoever died from a high fall (after climbing a tree to rescue a sloth, was this in the Rain Forest) traveled for less time than the person who died from Food Poisoning.
Valiant Jack traveled for 2 months less than the explorer eaten by a crocodile. Adventuresome Sam traveled for 1 month less than the explorer who went to the Frozen Wasteland. The explorer who went 'Sailing the Oceans' was away for the longest before succumbing to his fate. The explorer who died after 3 months didn't travel to the Long Lost City.
Barry Brave did not die of Food Poisoning. Courageous Fred traveled for 3 months more than Daring Derek. The explorer who visited the Dirty Desert traveled for 6 months more than the person who died of Hypothermia after they tripped over a seal and plunged head first through thin ice.
Name. Place. Cause of Death. Months Traveling.
