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Probability is Global + Are (a,b) and [a,b] the same length?

 Hey everyone! Welcome back to my blog! 


The number of diagonals are \binom {12}{2}-30=36

Today we have another topic which I am very scared of, PROBABILITY! I think to overcome this fear, I should start from basics. So, I refer AOPS's introduction to counting and probability. I did till chapter seven and thought to do the rest problems in this blog as notes.

The book's pirated copy is available in zlib btw ( I don't think so I should have said this, but I think it's fine). I did the harder problems and wrote the solutions in Xournal. If you want to see them, here is the pdf.

Also both the pdf and following post has amc/ioqm type problems! So enjoy!!

Problem 1: n coins are flipped simultaneously flipped. The probability that at most one of them shows tails is \frac{3}{16}. Find n.

Solution: We have \frac{n+1}{2^n}=\frac{3}{16}\implies 16(n+1)=3\cdot 2^n \implies n\ge 4.   But the RHS grows very fast, so n=5 is the only solution.

Intersection of indepent events: 
If A and B are possible outcomes for two independent events, then P(A \text{ and } B)=P(A)\times P(B)

Problem 2: The Grunters and the Screamers are playing for the Grand Championship, which is a best of 7 series. The first team to win 4 games wins the Championship. Each team has a \frac{3}{4} probability of winning any individual game. If the Grand Championship series lasts exactly 6 games, what is the probability that the Grunters win?

Solution: Note that if game lasted till 6 game then last game was won by G and so there are \binom{5}{3}=10 such winning games.

The probability of Grunters winning in 6 games is
P(\text{ Grunters win in $6$ games })= P(GGSGSG)+P(GSSGGG)+\dots ( since they are mutually exclusive, we can add).

Now, P(GGSGSG)= P(G)\times P(G)\times P(S)\times P(G)\times P(S)\times P(G)
= P(G)^4\times P(S)^2=\frac{3}{4}^4\times \frac{1}{4}^2= \frac{3^4}{4^6}. ( As they are independent events).

And for any winning game, we have probability \frac{3^4}{4^6}.

So, P(\text{ Grunters win in $6$ games })= P(GGSGSG)+P(GSSGGG)+\dots
= \binom{5}{3}\times \frac{3^4}{4^6}= \frac{405}{2048}

Problem 3: The probability of getting rain on any given day in June in Capital City is \frac{1}{10}. What is the probability that it rains on at most 2 days in June? 

Solution: Note that P( \text{ rains on almost 2 days in June}) = P( \text{rains on 0 day })+ P( \text{rains on 1 day })+ P( \text{rains on 2 days })$ 

= \frac{9}{10}^{30}+\frac{9}{10}^{29}\cdot \frac{1}{10}+\frac{9}{10}^{28}\cdot \frac{1}{10}^2=\frac{9^{30}+9^{29}+9^{28}}{10^{30}}

Probability with Dependant Events:

Problem 4: A bag has 4 red and 6 blue marbles. A marble is selected and not replaced, then a second is selected. What is the probability that both are the same colour?

Solution: P( \text{ both red })+P( \text{ both blue})=\frac{4}{10}\times \frac{3}{9} +\frac{6}{10}\times \frac{5}{9}

Problem 5: Sheila has been invited to a picnic tomorrow. The picnic will occur, rain or shine. If it rains, there is a 20%  probability that Sheila will decide to go, but if it is sunny, there is an 80% probability that Sheila will decide to go. The forecast for tomorrow states that there is a 40% chance of rain. What is the probability that Sheila will attend the picnic?

Solution: The required probability is P(\text{ it rains and she goes })+P(\text{ it doesn't rain and she goes }) 
=\frac{40}{100}\cdot \frac{20}{100}+\frac{60}{100}\cdot \frac{80}{100}= \frac{56}{100}.

The shooting stars: ( it is a series of two problems)

Problem 6: On any given night, Becky has a 60% chance of seeing a shooting star in any given hour. If Becky watches the sky for two hours, what is the probability that becky sees a shooting star?

Solution: Note that P(\text{ Becky sees a shooting star })=1- P(\text{ Becky doesn't see a shooting star }) 
1-\frac{40}{100}\cdot \frac{40}{100}=1-\frac{16}{100}= \frac{84}{100}

Problem 7: On Saturday night, Becky again goes stargazing. This time, conditions are better and there's 80% chance that she will see a shooting star in any given hour. We assume that the probability of seeing a shooting star is uniform for the entire hour. What is the probability that becky will see a shooting star in the first 15 minutes?

Solution: Let p be the probability of seeing a shooting star in 15 minutes. Since it's uniform, we get p as the probability of seeing a shooting star in 2nd 15 minute, and so on.
So P(\text{ no shooting star seen in 1 hr })= 

P(\text{ no shooting star is 1st 15 mins })\times P(\text{ no shooting star is 2nd 15 mins }) \times P(\text{ no shooting star is 3rd 15 mins })\times P(\text{ no shooting star is 4th 15 mins })
\implies \frac{20}{100}=(1-p)^4\implies p=1-\frac 15^{1/4}

Problem 8: The numbers 1 through 8 are arranged to form an eight digit number which is a multiple of 5. What is the probability that it is greater than 60,000,000?

Solution: Our last digit is always 5. So the possible numbers which can be in first digit is 1,2,3,4,6,7,8. For the number to be greater than 60,000,000 we want first digit to be 6,7,8.

So the probability is \frac{3}{7}.

Problem 9: What is the probability that a random rearrangement of the letters in the word "Mathematics" will begin with the letter "Math" ?

Solution:  Total possibilities = \frac{11!}{2!2!2!}
Number of letter's with 'MATH' = 7!

P=\frac{8}{11\times10\times 9\times 8}=\frac{1}{990}

Another way is, prob of M getting choose as first letter is \frac{2}{11}, A as second letter is \frac{2}{10} and so on.

Problem 10: A 2005 \times 2005 square consists of (2005)^2 unit squares. The middle square of each row is shaded. If a rectangle (of any size) is chosen at random, what is the probability that the rectangle includes a shaded square?

Solution: We do complimentary counting, first. So P( \text{ rectangle including a shaded square })
= 1- P(\text{ rectangle not including a shaded square })= 1- \frac{\text{ number of rectangles not including square}}{\binom{2006}{2}^2}
= 1 - 2\frac{\binom{1003}{2}\cdot \binom{2006}{2}}{ \binom{2006}{2}^2}
= 1- \frac{1002}{2005}=\frac{1003}{2005}

Geometric Probability: 

P(\text{ Whatever that freaking ugly problem wants })=\frac{\text{ dimension of good region }}{\text{ dimension of possible region }}

Problem 11: AC has length 5, and AB has length 4. A point P is selected randomly on the segment AC. What is the probability that P is closer to B than to A?

Solution: Introduce the midpoint M of AB. Then when P is between A and M it is closer to A than P else it's closer to B. So P= \frac{MB+BC}{AC}=\frac{3}{5}.

Problem 12: A real number x is selected randomly such that 0\le x\le 3. What is the probability that |x-1|\le \frac{1}{2}?

Solution:  Possible range for x is [0.5,1.5], which has range length 1. So P(|x-1|\le \frac{1}{2}) = \frac{1}{3}.

Problem 13: A real number x is chosen at random such that 0<x<100. What is the probability that [\sqrt{x}] is even?

Solution: Since [\sqrt{x}] is even, So x \in \{ (0,1), [4,9), [16,25), [36, 49), [ 64, 81)\} 

So \text{possible length is }= 1+5+ 9+13+17 =45

So the probability is \frac{45}{100}.


Okay wait... what!!!!!!! Doing the above two questions, I was like, wait. Is the length of (a,b),[a,b),(a,b], [a,b] same? 

And turns out it is! Actually, by definition, the length of the interval is b-a. Read more here!

Problem 14: Let CD be a line segment of length 6. A point P is chosen at CD. What is the probability that the distance from P to C is smaller than the square of the distance from P to D?

Solution: Let the distance between CP be d\implies d\le (6-d)^2\implies d\le 4.
So P= \boxed{\frac{4}{6}}.

Problem 15: Suppose two number x and y are each chosen such that 0<x<1 and 0<y<1. What is the probability that x+y>\frac{3}{2}?

Solution: Note that both x,y are greater than 1/2.  So the probability is 1/2.

I think the following problem is just so black magic. Not my sol, but I found it so amazing :O

Problem 16: My friend and I are hoping to meet for lunch. We will each arrive at our favourite restaurant at a random time between noon and 1 pm, stay for 15 mins, then leave. What is the probability that we will meet each other while at the restaurant?

Solution: We make a graph, and black magic happens.

Problem 17: A point P is randomly chosen in the interior of the right triangle ABC. What is the probability [PBC]\le \frac 12 [ABC]?

Solution: [ABC]=\frac 12 AC\times BC. And so the distance between P and BC must be \le AC/2. 

So the probability is \boxed{ \frac 34}.

Then we have a well know problem :P which was in 3b1b :P

Problem 18: Three points are selected randomly on the circumference of a circle. What is the probability that the triangle formed by these three points contains the centre of the circle?




Anyways, that's it for today's probability! We actually have an integer poly lecture, so I will go and revise that! See ya! Oh and I hope, if I am not tired, then I will post GT notes!

Sunaina 💜

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