A fair coin is tossed 11 times. What is the probability that only the first two tosses will yield heads ?

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: 21/38

Explanation:

n(S) =  C 2 20 = 190 

n(E) =  C 2 15 = 105 

Therefore, P(E) = 105/190 = 21/38

Correct Answer: 9/13

Explanation:

There are 13 spades ( including one king). Besides there are 3 more kings in remaining 3 suits

 

Thus   n(E) = 13 + 3 = 16

 

Hence  n E ¯ = 52 - 16 = 36 

  

Therefore,  P E = 36 52 = 9 13

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.