Solutions Data Analysis

Q.1 – Q.2 ,   Q.3 – Q.4  Q.5  ,  Q.6 ,  Q.7  , Q.9 , Q. 10  , Q. 11  Q.12 ,   Q.13 , Q. 14 , Q. 15   Q.16 to Q. 19

Answer 11.

All positive 2 digit integers   are 10 to 99 ; total 90 integers

4 in tens place will be  40 , 41 , 42 , …………49  ; total 10 integers

(a). 

the probability of a  4 in tens place  P(M) = 10/90

                                                                 P(M)   =1/9

 (b).

at least one 4 at unit place

4 at units place will be ; 14, 24, 34, 44, 54, 64, 74, 84, 94

Total 9 options available for 4 at unit place

        Probability of 4 at unit place P(N)  = 9/90 = 1/10

So , Probability of at least one 4 in tens place or the  unit place

Here both the event are independent to each other

P (M∪N) = P (M) + P (N) – P(M)P(N)

=   10/90 + 9/90 – (10/90) x (9/90)

= 1/9 + 1/10  – (1/9 x 1/10)

= 19/90 – 1/90

= 18/90

                                   P (M∪N)   = 1/5

( c ).

No  4  at unit or tens place

Probability of not having ‘4’ at unit or tens place  =  1 – 1 / 5

                                                                      = 4/5

Answer 12

Probability of defective parts = P(M) = 2/10

= 1/5

( a ). 

 Probability of choosing one part which is not defective = 1 –  P ( M )

= 1 – 1/5

= 4/5

Or

Out of 10 there are 2 defective , means  10 – 2 = 8 are not defective

So, Probability of choosing one part which is not defective = 8/10

                                                           = 4/5

( b )

Probability of taking part which is defective  P (  M ) = 2/ 10 = 1 / 5

Now the part is not replaced , so there are only 10 – 1  = 9 parts are available for next choice and there are only 2 defective parts

So, Probability of taking second part which is defective P (N) = 1/9

P(M)  and P (N ) = 1/5 x 1 /9

= 1/45

                                                                              

Q.1 – Q.2 ,   Q.3 – Q.4  Q.5  ,  Q.6 ,  Q.7  , Q.9 , Q. 10  , Q. 11  Q.12 ,   Q.13 , Q. 14 , Q. 15   Q.16 to Q. 19

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