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Probability Basics

Probability is the likelihood or chance of an event occurring.
Probability = the number of ways of achieving success/ the total number of possible outcomes

For example, the probability of flipping a coin and it being heads is ½, because there is 1 way of getting a head and the total number of possible outcomes is 2 (a head or tail). We write P(heads) = ½ .

The probability of something which is certain to happen is 1.
The probability of something which is impossible to happen is 0.
The probability of something not happening is 1 minus the probability that it will happen.

Illustrations

Example
There are 6 beads in a bag, 3 are red, 2 are yellow and 1 is blue. What is the probability of picking a yellow?

The probability is the number of yellows in the bag divided by the total number of balls, i.e. 2/6 = 1/3.

Example
There is a bag full of coloured balls, red, blue, green and orange. Balls are picked out and replaced. John did this 1000 times and obtained the following results:
Number of blue balls picked out: 300
Number of red balls: 200
Number of green balls: 450
Number of orange balls: 50

a) What is the probability of picking a green ball?
b) If there are 100 balls in the bag, how many of them are likely to be green?

a) For every 1000 balls picked out, 450 are green. Therefore P(green) = 450/1000 = 0.45

b) The experiment suggests that 450 out of 1000 balls are green. Therefore, out of 100 balls, 45 are green (using ratios).

Example
Find a chance of throwing more than 15 in one throw with 3 dice.
1) 1/54 2)17/216 3)5/108 4) CANNOT BE DETERMINED
Probability = the number of ways of achieving success
the total number of possible outcomes
Total no of possible outcomes with 1 dice = 6
Total no of possible outcomes with 3 dice = 6x6x6 =216
Now the denominator is fixed. We have to find out the numerator.
Possible favourable outcomes are 6,6,6 ; 6,6,5 ; 6,5,5 and 6,6,4.
For 6,6,6: no of ways =1
For 6,6,5 no of ways = 3
For 6,5,5 no of ways = 6,5,5 or 5,6,5 or 5,5,6 so 3.
For 6,6,4 no of ways = 3
Total no of favourable ways = 3+ +3+1 =10
Thus Ans =10/216 = 5/108 ie 3rd option.

Some more Basic Fundas

In mathematics a probability of an event, A is represented by a real number in the range from 0 to 1 and written as P(A), p(A) or Pr(A). An impossible event has a probability of 0, and a certain event has a probability of 1.
The opposite or complement of an event A is the event [not A] (that is, the event of A not occurring); its probability is given by P(not A) = 1 - P(A).

There are two very common terms in Probability : Independent and Mutually Exclusive.

For example, when drawing a single card at random from a regular deck of cards, the chance of getting a heart or a face card (J,Q,K) (or one that is both) is , because of the 52 cards of a deck 13 are hearts, 12 are face cards, and 3 are both: here the possibilities included in the "3 that are both" are included in each of the "13 hearts" and the "12 face cards" but should only be counted once.

Conditional probability is the probability of some event A, given the occurrence of some other event B. Conditional probability is written P(A|B), and is read "the probability of A, given B". It is defined by

If P(B) = 0 then is undefined.
For Independent events , P(B/A)=P(B)
And P(A/B)=P(A)
Bayer’s Theorem

=P(B/A) x P(A)/P(B)

Possibility Spaces

When working out what the probability of two things happening is, a probability/ possibility space can be drawn.

Example
if you throw two dice, what is the probability that you will get: a) 8, b) 9, c) either 8or9?
a) The black blobs indicate the ways of getting 8 (a 2 and a 6, a 3 and a 5, ...). There are 5 different ways. The probability space shows us that when throwing 2 dice, there are 36 different possibilities (36 squares). With 5 of these possibilities, you will get 8. Therefore P(8) = 5/36 .
b) The red blobs indicate the ways of getting 9. There are four ways, therefore P(9) = 4/36 = 1/9.
c) You will get an 8 or 9 in any of the 'blobbed' squares. There are 9 altogether, so P(8 or 9) = 9/36 = 1/4.

Probability Trees

Another way of representing 2 or more events is on a probability tree.

Example
There are 3 balls in a bag: red, yellow and blue. One ball is picked out, and not replaced, and then another ball is picked out.

The first ball can be red, yellow or blue. The probability is 1/3 for each of these. If a red ball is picked out, there will be two balls left, a yellow and blue. The probability the second ball will be yellow is 1/2 and the probability the second ball will be blue is 1/2. The same logic can be applied to the cases of when a yellow or blue ball is picked out first.

In this example, the question states that the ball is not replaced. If it was, the probability of picking a red ball (etc.) the second time will be the same as the first (i.e. 1/3).

In the above example, the probability of picking a red first is 1/3 and a yellow second is 1/2. The probability that a red AND then a yellow will be picked is 1/3 × 1/2 = 1/6 (this is shown at the end of the branch). The rule is:
• If two events A and B are independent (this means that one event does not depend on the other), then the probability of both A and B occurring is found by multiplying the probability of A occurring by the probability of B occurring.
The probability of picking a red OR yellow first is 1/3 + 1/3 = 2/3. The rule is:
• If we have two events A and B and it isn't possible for both events to occur, then the probability of A or B occuring is the probability of A occurring + the probability of B occurring.

On a probability tree, when moving from left to right we multiply and when moving down we add.

Example
What is the probability of getting a yellow and a red in any order?
This is the same as: what is the probability of getting a yellow AND a red OR a red AND a yellow.
P(yellow and red) = 1/3 × 1/2 = 1/6
P(red and yellow) = 1/3 × 1/2 = 1/6
P(yellow and red or red and yellow) = 1/6 + 1/6 = 1/3

Example

In a Shooting Competition, the probability of hitting the target by A is 2/5, by B is 2/3 and by C is 3/5. If all of them fire independently at the same target, then find the probability that only one of them will hit the target.

P(A)=2/5 P(A*) = 3/5

P(B)=2/3 P(B*) = 1/3

P(C)=3/5 P(C*) = 2/5

Probability that only one of them hits the target
= Probability that A hits the Target but not B and C
+ Probability that B hits the Target but not A and C
+ Probability that C hits the Target but not B and A

= P(A @ B* @ C*) + P(B @ C* @ A*) + P(C @ A* @ B*)

Where @ represents intersection symbol.

= 2/5 x 1/3 x 2/5 + 2/3 x 3/5 x 2/5 + 3/5 x 3/5 x 1/3
= 1/3 Ans

Example

In a fruit Basket 40% of the fruits are mangoes and rest are apples.
If 25% of the mangoes are ripe and 10% of the apples are ripe, find the probability that a Ripe fruit randomly selected is a Mango.

Suppose there are 100 fruits.
No of mangoes = 40 and apples = 60
No of ripe mangoes = 10 and ripe apples = 6

So here we have multiple events taking place. One is the selection of fruit : Mango or Apple. Other is the Selection of Type : Unripe or Ripe.

Let say Selection of Mango as event X and selection of ripe fruit as event Y .

P(Y)= Prob of selection of a ripe fruit = 16/100

P(X@Y) = Probabilty of selection of a Ripe Mango = 10/100

But what is asked is P(X/Y) = P(X@Y)/P(Y) = 10/100 divided by 16/100 = 10/16

= 5/8.

Brain Teaser

Bill and Ben takes turns tossing a coin. Whoever gets the head
first is the winner. If bill has the first toss, what is the
probability that he will win?

Ans : 2/3

Solution : Prob that A wins = prob of a head in 1st turn
+ prob of a head in 3rd turn
+ prob of a head in 5th turn
+ prob of a head in 7th turn
+ ….so on upto infinity

= 1/2 + (1/2)^3 + (1/2)^5 + …..

= ½ [ 1/ (1-{(1/2)^2})]

= 2/3.

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