Guide Solution Data Analysis – MS

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 13

( a ).

Probability of selecting a junior = P ( M ) =  19/ 40

So, Probability of not selecting a junior  = 1 – P ( M)

=  1 – 19 / 40

= 21/40

        OR

There are  19 juniors  that means  40 – 19 = 21 are not juniors

Probability of not selecting a junior  = 21/40

 ( b ) .

 Probability of a female or a Sophomores

There are total 22 females,

So,   Probability of  selecting  a female = P (N)= 22/40

There are total 16 Sophomores and out of this, there are 10 females which we have already considered, hence here we consider males only

Probability of selecting a Sophomores  = P ( L ) =  6/40

As both event are mutually exclusive  , apply the formula

Probability of selecting a female or a  Sophomores = P (N ∪ L)

P (N∪L) = P (N) + P (L)

P (N ∪ L) = 22/40 + 6/40

= 28/40

= 7/10

( c ).

Probability of  selecting a male sophomores = P (M ) = 6/40

Probability of selecting a female senior = P ( N )  =  3/40

Both are mutually exclusive events

Probability of selecting a male sophomores or a female senior ;

P ( M ∪  N )  = P ( M ) + P ( N )

=   6 /40 + 3 /40

= 9/40

Answer 14

(a).

A and B are mutually exclusive events so,

P ( A∪ B ) =  P ( A ) + P (B )

0.6  = 0.2  + P ( B )

P ( B ) =  0.6 -0.2

P ( B ) = 0.4

(b).

C and D are independent events

P ( C ∪ D )  =  P ( C )  + P ( D )  – P ( C ) x P ( D )

0.6  = 0.5  + P (D)  – 0.5 P(D)

0.1  = 0.5 P ( D )

P ( D ) = 0.2

Answer 15

Probability that Lin  decode a message = P( M)  = 0.8

Probability that Mark decode a message = P ( N ) = 0.7

Both are independent events

(a).

Both will decode the message  = P ( M ∩ N ) =  P ( M ) x  P ( N )

=  0. 8 x  0. 7

P ( M ∩ N )= 0.56

( b ) .

At least one of them will decode the message

Probability of at least one of them will decode the message = P ( M ∪ N )

For independent events ;

P (M∪N) = P (M) + P (N) – P(M)P(N)

=   0.8 + 0. 7  – 0. 8 x 0. 7

= 1.5 – 0.56

= 0.94

( c ) .

Probability of neither of them decode the message = 1 – P(M ∪ N)

= 1 – 0.94

= 0.06

                                                                                                                        

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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