A card is drawn from a pack of 52 cards. The card is drawn at random. What is the probability that it is neither a spade nor a Jack?

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: 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: 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: (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: 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.

Correct Answer: 1/5

Explanation:

Number of applicants = 5
On a day, only 1 leave is approved.
Now favourable events =  1 of 5 applicants is approved
Probability that Laxmi priya's leave is granted = 1/5.

Correct Answer: 2/3

Explanation:

S = { 1, 2, 3, 4, .....18 } 

=> n(S) = 18

 

E1 = {2, 4, 6, 8, 10, 12, 14, 16, 18}

=> n(E1) = 9

 

E2 = {3, 6, 9, 12, 15, 18 }

=> n(E2) = 6

 

  E 3 = E 1 ? E 2 = { 6 , 12 , 18 } 

=> n(E3) = 3

 

? E = E 1 ? E 2 = E 1 + E 2 - E 3 

=> n(E) = 9 + 6 - 3 =12

 where E = { 2, 3, 4, 6, 8, 9, 10, 12, 12, 14, 15, 16, 18 }

 

? P ( E ) = n ( E ) n ( S ) = 12 18 = 2 3

Correct Answer: 55/221

Explanation:

n(S) =  C 2 52 = 1326

 

 Let  A = event of getting both red cards

 

and B = event of getting both queens

 

then  A ? B = event of getting two red queens

 

n(A) =  C 2 26 = 325,   n(B) =  C 2 4 = 6

 

  n ( A ? B ) = C 2 2 = 1

 

  ? P A = 325 1326 , P B = 6 1326

 

  P A ? B = 1 1326

 

?P ( both red or both queens) =  P A ? B

 

=  P A + P B - P A ? B = 325 1326 + 1 221 - 1 1326= 55 221