Probability

Probability Example

Probability Example 10

Probability Example 6-9

Probability Example 5

Probability Example

Two dice are thrown, what is the probability of getting the sum being 8 or the sum being 10?

Solution:

Number of sample points in throwing two dice at a time is n(S)=36 Let A= {the sum being 8}

\      A= {(6,2), (5,3) , (4,4), (3,5) , (2,6)}; P(A) =  5/36

B = {the sum being 10}

\     B = {(6,4), (5,5) (4,6)} ;                        P(B) =  3/36

A Ç B = { } ;  n(A Ç B) = 0

\     The two events are mutually exclusive

\     P(A È B) = P(A) + P(B)

5      3

=      --- + ---          

36    36

8      2

=   ----- = --

36     9

Solves Examples on Probability

Example 1:

Three coins are tossed simultaneously Find the probability that

(i) no head                               (ii) one head                                        (iii) two heads

(iv)     atleast two heads.             (v) atmost two heads appear.

Solution:

The sample space for the 3 coins is

S = { HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} ; n(S) = 8

(i)                   No head appear A = {TTT}; n(A) = 1

\P(A) = 1/8

(ii)                 One head appear B = {HTT, THT, TTH}; n (B) = 3

\P(B) = 3/8

(iii)                Two heads appear C = {HHT, HTH, THH}; n(C)=3

\P(C) = 3/8

(iv)                Atleast two heads appear

D = { HHT, HTH, THH, HHH}; n(D) = 4

                P (D)      4/8 =1/2               

(v)                 Atmost two heads appear E = { TTT, HTT, THT, TTH,HHT, HTH,THH} n(E) = 7

\P(E) =  7/8

Example 2:

When two dice are thrown, find the probability of getting doublets (Same number on

both dice)

Solution:

When two dice are thrown, the number of points in the sample space is n(S) = 36 Getting doublets: A = {(1,1) , (2,2) , (3,3) , (4,4) , (5,5) , (6,6)}

\P(A) =   6/36 = 1/6

Example 3:

A card is drawn at random from a well shuffled pack of 52 cards. What is the probability that it is (i) an ace (ii) a diamond card

Solution:

We know that the Pack contains 52 cards \ n(S) = 52

(i)    There are 4 aces in a pack. n(A) = 4

\P(A) =  4/52 =1/13

(ii)     There are 13 diamonds in a pack \ n(B) = 13

\P(B) =  13/52 = 1/4

Example 4:

A ball is drawn at random from a box containing 5 green, 6 red, and 4 yellow balls. Determine the probability that the ball drawn is (i) green (ii) Red (iii) yellow (iv) Green or Red (v) not yellow.

Solution:

Total number of balls in the box = 5 + 6 + 4 = 15 balls

(i)	Probability of drawing a green ball = 5/15 =1/3
(ii)	Probability of drawing a red ball= 6/15 = 2/5
(iii) (iv)	Probability of drawing a yellow ball=4/15 Probability of drawing a Green or a Red ball

               5/15 + 6/15 = 11/15

Basic principles of Permutation and Combination

Factorial

The consecutive product of first n natural numbers is known as factorial n and is denoted as n! or Ðn

That is n! = 1 × 2 × 3 × 4 × 5 × ... × n 3! = 3 × 2 × 1

4! = 4 × 3 × 2 × 1

5! = 5 × 4 × 3 × 2 × 1

Also     5! = 5 × ( 4 × 3 × 2 × 1 ) = 5 × ( 4! )

Therefore this can be algebraically written as n! = n × (n – 1)! Note that 1! = 1 and 0! = 1.

Permutation: 

Permutation means arrangement of things in different ways. Out of three things A, B, C taking two at a time, we can arrange them in the following manner.

A B                              B A

A C                              C A

B C                              C B

Here  we  find  6  arrangements.  In  these  arrangements  order  of  arrangement  is

considered. The arrangement AB and the other arrangement BA are different.

The number of arrangements of the above is given as the number of permutations of 3 things taken 2 at a time which gives the value 6. This is written symbolically, 3P2 = 6

Thus the number of arrangements that can be made out of n things taken r at a time is known as the number of permutation of n things taken r at a time and is denoted as nPr.

The expansion of nPr is given below:

nPr = n(n-1)(n-2) ……………[n – ( r – 1)]

The same can be written in factorial notation as follows:

nPr =

n! (n - r)!

For example, to find 10P3 we write this as follows:

10P3      = 10 (10 – 1) (10 – 2)

= 10 × 9 × 8

= 720

[To find 10P3, Start with 10, write the product of 3 consecutive natural numbers in the descending order]

Simplifying 10P3 using factorial notation:

10 P3

    10!          10´ 9 ´8 ´ 7 ´ 6 ´ 5 ´ 4 ´ 3 ´ 2 ´ 1

=                =

(10 - 3)!               7 ´ 6 ´ 5 ´ 4 ´ 3 ´ 2 ´ 1

= 10 ´ 9´ 8

= 720

Note that nP0 = 1, nP1 = n, nPn = n!

Combinations:

A combination is a selection of objects without considering the order of arrangements.

For example, out of three things A, B, C we have to select two things at a time. This can be selected in three different ways as follows:

A B                  A C                  B C

Here the selection of the object A B and B A are one and the same. Hence the order of arrangement is not considered in combination. Here the number of combinations from 3 different things taken 2 at a time is 3.

This is written symbolically 3C2 = 3

Thus the number of combination of n different things, taken r at a time is given by

nCr = n Pr

r!

Or  nCr =

n!

(n - r)! r!

Note that nC0 = 1,     nC1 = n,        nCn = 1

Find      10C3 ,     =  10 ´ 9 ´8/ 3x2 = 120

   3

Find 8C4  ,                  8C4

(8 ´ 7 ´ 6 ´ 5) / 1x2x3x4 = 70

                    

Combinations:

A combination is a selection of objects without considering the order of arrangements.

For example, out of three things A, B, C we have to select two things at a time. This can be selected in three different ways as follows:

A B                  A C                  B C

Here the selection of the object A B and B A are one and the same. Hence the order of arrangement is not considered in combination. Here the number of combinations from 3 different things taken 2 at a time is 3.

This is written symbolically 3C2 = 3

Thus the number of combination of n different things, taken r at a time is given by

nCr = n Pr

r!

Or  nCr =

n!

(n - r)! r!

Note that nC0 = 1,     nC1 = n,        nCn = 1

Find      C ,       C

= 10 P3

10 ´ 9 ´8

=

= 120

10    3

10    3

              

3!        1´ 2 ´ 3

Find 8C4  ,                  8C4

8 ´ 7 ´ 6 ´ 5

=                    = 70

1 ´ 2 ´ 3 ´ 4

[ To find 8C4 : In the numerator, first write the product of 4 natural numbers starting

with 8 in descending order and in the denominator write the factorial 4 and then simplify.]

Compare 10C8 and 10C2

10 C8

10 C2

10 ´ 9 ´ 8 ´ 7 ´ 6 ´ 5 ´ 4 ´ 3    10 ´ 9

=                                             =            = 45

1 ´ 2 ´ 3 ´ 4 ´ 5 ´ 6 ´ 7 ´ 8      1 ´ 2

10 ´ 9

=            = 45

1 ´ 2

From the above, we find 10C8 = 10C2

This can be got by the following method also:

10C8 = 10C(10 – 8) = 10C2

This method is very useful, when the difference between n and r is very high in nCr.

This property of the combination is written as nCr = nC(n-r).

To find 200C198 we can use the above formula as follows:

nCr      =       n!  / (n - r)! r!

200 ´ 199/2                  = 19900.