PROBABILITY

Ans .

1/2

1. Explanation :

   Here s={H,T} and E={H}.

   P(E)=n(E)/n(S)=1/2

Ans .

3/4

1. Explanation :

   Here S={HH,HT,TH,TT}

   Let Ee=event of getting one head

   E={TT,HT,TH}

   P(E)=n(E)/n(S)=3/4

Ans .

1/3

1. Explanation :

   Here S={1,2,3,4,5,6}

   Let E be the event of getting the multiple of 3

   then ,E={3,6}

   P(E)=n(E)/n(S)=2/6=1/3

Ans .

5/12

1. Explanation :

   Here n(S)=(6*6)=36

   let E=event of getting a total more than 7

   ={(2,6),(3,5),(3,6),(4,4),(4,5),(4,6),(5,3),(5,4),(5,5),(5,6),(6,2),(6,3),(6,4),(6,5),(6,6)}

   P(E)=n(E)/n(S)=15/36=5/12.

Ans .

7/15

1. Explanation :

   .let S be the sample space

   Then n(S)=no of ways of drawing 2 balls out of (6+4)=(\(_{2}^{10}\textrm{C}\)=(10*9)/(2*1)=45

   Let E=event of getting both balls of same colour

   Then n(E)=no of ways(2 balls out of six) or(2 balls out of 4)

   =((\(_{2}^{6}\textrm{C}\)+(\(_{2}^{4}\textrm{C}\))=(6*5)/(2*1)+(4*3)/(2*1)=15+6=21

   P(E)=n(E)/n(S)=21/45=7/15

Ans .

7/18

1. Explanation :

   Clearly n(S)=6*6=36

   Let E be the event that the sum of the numbers on the two faces is divided by

   4 or 6.Then

   E={(1,3),(1,5),(2,2),(2,4),(2,6),(3,1),(3,3),(3,5),(4,2),(4,4),(5,1),(5,3),(6,2),(6,6)}

   n(E)=14.

   Hence p(e)=n(e)/n(s)=14/36=7/18

Ans .

55/221

1. Explanation :

   We have n(s)=52c2=(52*51)/(2*1)=1326.

   Let A=event of getting both black cards

   B=event of getting both queens

   A\(\cap\)B=event of getting queen of black cards

   n(A)= (\(_{2}^{26}\textrm{C}\)=(26*25)/(2*1)=325,

   n(B)= (\(_{2}^{4}\textrm{C}\)=(4*3)/(2*1)=6 and

   n(A\(\cap\)B)= (\(_{2}^{2}\textrm{C}\)=1

   P(A)=n(A)/n(S)=325/1326;

   P(B)=n(B)/n(S)=6/1326 and

   P(A\(\cap\)B)=n(A\(\cap\)B)/n(S)=1/1326

   P(A\(\cap\)B)=P(A)+P(B)-P(A\(\cap\)B)=(325+6-1/1326)=330/1326=55/221.