PROBLEM BASED ON P(A)=n(A)/n(B)

 

PROBLEM 1.

Find the probability that exactly one head appears in a single throw of a fair coin

Solution: Here throw a single coin get a sample space

 $ S = \lbrace H,T \rbrace $
 $ n(S) = 2  $

Let A be a event of getting exactly one head in a single throw of a fair coin

 $ A  = \lbrace H \rbrace $
 $n(A) = 1 $

 $ P(A) = \frac{n(A)}{n(S)} $
$ \therefore  P(A) = \frac{1}{2}   $ 

 

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PROBLEM 2 

A lot of integrated circuit chips consists of 10 good,4 with minor defects and 2  with major defects.Two chips are randomly chosen from the lot.What is the probability that at-least one chips is good?

Solution: Here
  

Integrated circuit chips   Good    Minor defect   Major defect  Total
                                        10               4                   2                  16

Let S be chosen two chips randomly selected from the lot of integrated circuit.

$ n(S) = n C_{r}$   use \enspace  combination \enspace property
       $ = 16 C_{2} $  [n=16  is  total  chips r=2   is  select  two  chips randomly]
       $ = \frac{16!}{(16-2)! 2!}$  $ [n C_{r}=\frac{n!}{(n-r)! r!}]$ 
       $ =  \frac{14!\times 15 \times 16}{14! \times 2!}$  $ [n!=1\times2\times \dots (n-1)\times n ] $
      $ = \frac{15 \times 16}{2 \times 1}$  [cancel   common  term]
      $ = 15 \times 8 $ 

$ n(S) = 120$ [Another  way  use    calculator   fx-991MS]

Lex X(0,1,2) choose  are good  chips

 

X=0 $ \implies$  P(X=0) =$ \frac{10C_0 \times 6C_2 }{12C_2}$ [X=0 means   selected   no   good  chips ]
                                    $ = \frac{1 \times 15 }{120} $           [$10C_0=1,6C_2=15,12C_2=120 $ use   above    $ nC_r $  step ]
                                   $ =  \frac{15}{120}$    [use  calculator    fx-991MS ]
                                    $ = \frac{1}{8}$     [cancel  common  term ]

X=1  $ \implies $  P(X=1) = $ \frac{10C_1 \times 6C_1 }{12C_2} $ [X=1 selected  one  good  chips ] 

              $ = \frac{10 \times 6 }{120}$ [$ 10C_1=10,6C_1=6,12C_2=120 $  use  above  $ nC_r  $   step ]                                  

                                    $ =  \frac{60}{120} $ [use   calculator   fx-991MS ]  
                                    $ = \frac{1}{2} $ [cancel   common  term ]

X=2 $ \implies $  P(X=2)= $  \frac{10C_2 \times 6C_0 }{12C_2}$ [X=2 means   selected   two  good  chips ]
                                  $ = \frac{45 \times 1 }{120}$ [$ 10C_2=45,6C_0=1,12C_2=120 $  use   above   $ nC_r $   step ]
                                 $ =  \frac{45}{120}$ [use calculator    fx-991MS ]
$ =  \frac{3}{8}$ [cancel   common  term ]

The Probability Functions


$ X=x_i$               0               1              2 

$ P(X=x_i)$       $\frac{1}{8}$              $\frac{1}{2}$           $\frac{3}{8}$ 


Let X be a probability of atleast one good chips are good

P(atleast   one    is  good) = P($ X \geq 1$)
                                          = P(X=1)+P(X=2)                                        

                                        $=\frac{1}{2}+\frac{3}{8} $ [by  above  table  ]
                                       $ = \frac{4+3}{8}$
                                       $ = \frac{7}{8}$

 

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