One card is drawn from a pack of 52 cards , each of the 52 cards being equally likely to be drawn. Find the probability that the card drawn is neither a spade nor a king.

Correct Answer: (1/2)^11

Explanation:

Probability of occurrence of an event, 

P(E) = Number of favorable outcomes/Numeber of possible outcomes = n(E)/n(S)

? Probability of getting head in one coin = ½, 

? Probability of not getting head in one coin = 1- ½ = ½, 

Hence, 

All the 11 tosses are independent of each other.

? Required probability of getting only 2 times heads = 1 2 2 × 1 2 9 = 1 2 11

Correct Answer: 1/30

Correct Answer: 1/2

Explanation:

P(odd) = P (even) = 1 2 1(because there are 50 odd and 50 even numbers)

 

Sum or the three numbers can be odd only under the following 4 scenarios:

 

Odd + Odd + Odd =  1 2 * 1 2 * 1 2=  1 8

 

Odd + Even + Even =  1 2 * 1 2 * 1 2= 1 8

 

Even + Odd + Even =  1 2 * 1 2 * 1 2= 1 8

 

Even + Even + Odd =  1 2 * 1 2 * 1 2 =  1 8

 

Other combinations of odd and even will give even numbers. 

 

Adding up the 4 scenarios above:

 

=  1 8+  1 8+ 1 8+  1 8 =  4 8 =  1 2

Correct Answer: 1/2

Explanation:

Here, s={H,T} and E={H}
P(E) = n(E)/n(S) = 1/2

Correct Answer: 0.043

Explanation:

 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 ) = 5 P 5 2 ! * 2 ! ( A , I a r e 2 t i m e s ) * 6 P 6 2 ! ( T i s 2 t i m e s ) = 21600

 

n ( S ) = 11 ! 2 ! * 2 ! * 2 ! ( A , I a r e 2 t i m e s ) = 4989600

 

21600 4989600 = 0 . 043

Correct Answer: 39/100

Explanation:

P ( neither A nor B) =  P A a n d B  

 

=   P A ? B=   P A ? B=  1 - P A ? B  

 

=  1 - 61 100 = 39 100

Correct Answer: 1/221

Explanation:

 Let S be the sample space

 

Then, n(S) =  52 C 2= 1326.

 

Let E = event of getting 2 kings out of 4.

 

n(E)  =  52 * 51 2 * 1= 6.

  4 C 2

=> 4 * 3 2 * 1

P ( E ) = n ( E ) n ( S ) = 6 1326 = 1 221

Correct Answer: 9/13

Explanation:

There are 13 spade and 3 more jack Probability of getting spade or a jack: =13+352=1652=413   So probability of getting neither spade nor a jack: =1?413 = 9/13

Correct Answer: 3/4

Explanation:

Let S be the sample space.
Here n(S)= 2 3 = 8
Let E be the event of getting atmost two heads. Then,
n(E) = {(H,T,T), (T,H,T), (T,T,H), (H,H,T), (T,H,H), (H,T,H)}
Required probability = n(E)/n(S) = 6/8 = 3/4.

Correct Answer: 3/10

Explanation:

Here, S = {1, 2, 3, 4, ...., 19, 20}=> n(s) = 20
Let E = event of getting a multiple of 4 or 15
=multiples od 4 are {4, 8, 12, 16, 20}
And multiples of 15 means multiples of 3 and 5
= {3, 6 , 9, 12, 15, 18, 5, 10, 15, 20}.
= the common multiple is only (15).
=> E = n(E)= 6
Required Probability = P(E) = n(E)/n(S) = 6/20 = 3/10.