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.