A card is drawn from a pack of 52 cards. The probability of getting a queen of club or a

king of heart is:

Let A be event of selecting a number divisible by 3. B be the event of selecting a number divisible by 7.

  

 A = { 3, 6, 9, 12, 15, 18, 21, 24, 27, 30 }, so n(A)=10
 B = { 7, 14, 21, 28 }, n(B)= 4 

 

Since A and B are not mutually exclusive So :

Let A be the event of A will tell truth. B be the event of B tell truth

 P(A)=23,P(Ac)=1-P(A)=13

P(B)=45,P(Bc)=1-P(B)=15 

When both agree then they say true or they say false together, that is 

 A∩B or Ac∩Bc

Also these events will be mutually exclusive :

 P(A∩B)+P(Ac∩Bc)=P(A)P(B)P(C)=23*45+13*15=35

Let A be the event of driver A drive safely after consuming liquor. 

Let B be the event of driver B drive safely after consuming liquor. 

Let C be the event of driver C drive safely after consuming liquor. 

P(A)=2/5,P(B)=3/7,P(C)=3/4

The events A, B and C are independent . Therefore,

PA∩B∩C=P(A)P(B)P(C)=25*37*34=970

Therefore, The probability that all the drivers will drive home safely after consuming liquor is 9/70

Let A be the event of first person hitting the target,

P(A) =33+5=38 (odd in favour)

Let B be the event of Second person hitting a target.

P(B)=23+2=25 (odd against)

Since both events are independent and both will hit the target so,

P(A∩B)= P(A)P(B)=38×25=320

If a number ends with 5, 0 then the number will be divisible by 5
Here only 5 is present, end place will be fixed by 5 so, 

 Vowels are A I A I O, 

 C    A   S    T   I    G   A    T   I    O   N

(O) (E) (O) (E) (O) (E) (O) (E) (O) (E) (O)

So there are 5 even places in which five vowels can be arranged and in rest of 6 places 6 constants can be arranged as follows :

n(E)=5P52!*2!(A,I are 2 times)*6P62!(T is 2 times)=21600

n(S)=11!2!*2!*2!(A, I are 2 times)=4989600

216004989600=0.043