Probability 6

Mixing multiple probabilities

Given a box of 10 coins out of which 9 coins are fair and one coin has both sides heads, what is the probability of getting all heads on picking a coin from the bag and flipping it 5 times.

There are two outcomes possible if we pick a coin from the bag.
The first outcome would be that we pick a normal coin and its probability would be 9/10 --- (1).
The second outcome would be that we pick a coin having both sided heads, whose probability is 1/10 --- (2).

Now after this,
Given a normal coin what is the probability of getting all 5 heads is, as we know it to be (1/2)*(1/2)*(1/2)*(1/2)*(1/2) = 1/32 --- (3).

Given a both sided heads coin, the probability of getting all 5 heads is 1 --- (4).

From (1), (2), (3) & (4) we get the
P(getting all 5 heads) = (P(picking the normal coin)*P(getting all heads provided its normal coin))+(P(Picking an unfair coin)*P(getting all heads provided its unfair coin))

= (1/32)*(9/10) + 1*(1/10) = 9/320 + 1/10 = 41/320

Probability 5

What is the probability of getting a 7 when two dice is thrown.

The total number of outcomes throwing 2 dice would be 36 as shown in the figure below.
As you can see in the figure, the number of times we get 7 is 6.



Therefore, the P(7) = 6/36 = 1/6.

Probability 4

Example:
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The probability of me making a shot in basketball is 80%, I have 3 shots and we need 1 pt to win the match. What is the probability of we winning the match?

P(we winning) = P(me making atleast one shot) = 1 - P(me making no shots at all)

P(me making no shots at all) = 0.2 * 0.2 * 0.2 = 0.008 = 0.8%

P(we winning) = 1-0.008 = 99.2%

Probability 3

Arrangements [Permutations] meeting Probability.

1) Find the probability of getting only one heads in 5 flips

Outcomes are

TTTTH
TTTHT
TTHTT
THTTT
HTTTT

The probability of occurrence of each of these outcomes is 1/32.
Since there are 5 such arrangements the total probability becomes 5/32.



2) Find the probability of not getting exactly one heads in 5 flips

This is exactly equal to all the outcomes except the above which results to 1-5/32 = 27/32

Probability 2

Probability Tree

Probability Tree is a very good tool in finding the probability of an occurrence of an event provided -
1) The outcomes or circumstances of an event are very small in number
2) The experiment should not be repeated large number of times

Note:
The depth of the tree is the number of times the experiment is repeated.
Example: Flipping coins 5 times.

The width of the tree is the count of all possible outcomes of an event.
Example: H & T are the 2 outcome of flipping the coin.

Lets take an example and solve the probability using Probability Tree

Find the probability of two heads occuring on two consecutive flips of coins, i.e., P(HH)

From the above probability tree the total out comes or circumstances of flipping coin twice is 4. Out of the 4 outcomes, the total number of times we get HH is 1, so the P(HH) = 1/4.

Another way to look at it is the probability of first flip leading to heads is 1/2, and this remains the same even during the second flip because both heads and tails are equally probable and they are not mutually exclusive so it leads to 1/2 * 1/2 = 1/4.

Note we multiply the probabilities because the events are not mutually exclusive.

Similary P(1H, 1T) = 2/4 = 1/2

P(HHHHH) = 1/2*1/2*1/2*1/2*1/2 = 1/32

P(Out of 7 flips, heads should not come even once) = p(TTTTTTT) = 1/128

Probability 1

Probability
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Two ways of looking at probability are:

1) Repeated experiments possible [Frequency Test]
Perform an experiment repeatedly and identify wht percent of times the experiment has
behaved the way u want it to behave.
Example: Flipping a coin hundred times to know the probability of occurrence of heads.

2) Repeated experiments not possible
Based on the data that you have, how strongly you believe that the experiment will behave
the way you want it to behave.
Example: What are the chances that its going to rain today.


Definition of Probability
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P(x) = number of events in which x is true
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total number of events or all possible outcomes of an event or all possible circumstances of an event.

In case of flipping a coin, the total number of circumstances or outcome of flipping a coin is 2, it can be either heads or tails.

assuming that the occurence of all the events are equally probable, i.e no single event should occur more number of times than other events.


Examples:
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P(Occurence of Head on flipping a coin) = 1 / 2 = 50%

Math: Find the digit

Find the Digit

Let N be the largest integer such that N and 7N both have 100 digits.

What is the 50th digit of n?

Solution

If N is the largest number such that N and 7N both have 100 digits, then 7N must be the largest multiple of 7 that has 100 digits. We can therefore find N just by dividing 7 into 9999...99 (100 9's).

By simple long division, we find that N starts off with repeated blocks of 142857. The first, seventh, thirteenth, ..., forty-ninth digits must therefore be 1's, and so the fiftieth digit is a 4.

Math: Formula for Squares

Formula for Squares

Show that if x is a postive integer, and both 2x + 1 and 3x + 1 are squares, then x is a multiple of 40.

Solution

We work with the modulo classes of 5 and 8. Any square number must leave remainder 0, 1 or 4 when divided by 5 or 8:

n (mod5) n2 (mod 5) 0 0 1 1 2 4 3 4 4 1

n (mod8) n2 (mod 8) 0 0 1 1 2 4 3 1 4 0 5 1 6 4 7 1

Now lets examine the remainders when 2x + 1 and 3x + 1 are divided by 5 and 8:

x (mod5) 2x+1 (mod 5) 3x+1 (mod 5) 0 1 1 1 3 4 2 0 2 3 2 0 4 4 3

x (mod8) 2x+1 (mod 8) 3x+1 (mod 8) 0 1 1 1 3 4 2 5 7 3 7 2 4 1 5 5 3 0 6 5 3 7 7 6

From the table on the left, we see that it is only when x = 0 (mod 5) that both 2x + 1 and 3x + 1 have remainders 0, 1 or 4 mod 5 that allow both of them to be squares. Similarly from the second table, we must have x = 0 (mod 8).

Thus 5 and 8 both divide x, and so x is a multiple of 40.

The first few values of x with both 2x + 1 and 3x + 1 square are x = 0, 40, 3960, 388080, 38027920, 3726348120, ...