What is the probability of getting 53 Mondays in a leap year?

Correct Answer: 5/12

Explanation:

Clearly, n(S) = (6 x 6) = 36.

Let E = Event that the sum is a prime number.

Then E= { (1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4,3),(5, 2), (5, 6), (6, 1), (6, 5) }

n(E) = 15.

P(E) = n(E)/n(S) = 15/36 = 5/12.

Correct Answer: 9/20

Explanation:

Here, S = {1, 2, 3, 4, ...., 19, 20}.

Let E = event of getting a multiple of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}.

P(E) = n(E)/n(S) = 9/20.

Correct Answer: 2/7

Explanation:

Total number of outcomes possible, n(S) = 10 + 25 = 35

 

Total number of prizes, n(E) = 10

  P ( E ) = n ( E ) n ( S ) = 10 35 = 2 7

Correct Answer: 2/7

Correct Answer: 55/221

Explanation:

We have n(s) = 52 C 2 52 = 52*51/2*1= 1326. 

Let A = event of getting both black cards 

     B = event of getting both queens 

A?B = event of getting queen of black cards 

n(A) = 52 * 51 2 * 1 =  26 C 2 = 325, n(B)= 26 * 25 2 * 1= 4*3/2*1= 6  and  n(A?B) =  4 C 2 = 1 

P(A) = n(A)/n(S) = 325/1326;

P(B) = n(B)/n(S) = 6/1326 and 

P(A?B) = n(A?B)/n(S) = 1/1326 

P(A?B) = P(A) + P(B) - P(A?B) = (325+6-1) / 1326 = 330/1326 = 55/221

Correct Answer: 2/91

Explanation:

Let S be the sample space.

Then, n(S) = number of ways of drawing 3 balls out of 15 =  15 C 3  = 15 * 14 * 13 3 * 2 * 1= 455.  

Let E = event of getting all the 3 red balls.

 n(E) = 5 C 3 = 5 * 4 2 * 1 = 10.

 => P(E) = n(E)/n(S) = 10/455 = 2/91.

Correct Answer: 7/18

Explanation:

Clearly, n(S) = 6 x 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

Correct Answer: 19/90

Explanation:

A S S I S T A N T ? A A I N S S S T T

S T A T I S T I C S ? A C I I S S S T T T

Here N and C are not common and same letters can be A, I, S, T. Therefore

 Probability of choosing A =   2 C 1 9 C 1 × 1 C 1 10 C 1 = 1/45 

 Probability of choosing I =  1 9 C 1 × 2 C 1 10 C 1 = 1/45

Probability of choosing S =  3 C 1 9 C 1 × 3 C 1 10 C 1 = 1/10

Probability of choosing T =  2 C 1 9 C 1 × 3 C 1 10 C 1 = 1/15

Hence, Required probability =    1 45 + 1 45 + 1 10 + 1 15 = 19 90 

Correct Answer: 21/46

Correct Answer: 316/435

Explanation:

n s = C 2 30

 

 Let A be the event of getting two oranges and 

 

 B be the event of getting two non-defective fruits.

 

 and  A ? B be the event of getting two non-defective oranges

 

  ? P A = C 2 20 C 2 30 , P B = C 2 22 C 2 30 a n d P A ? B = C 2 15 C 2 30

 

  ? P A ? B = P A + P B - P A ? B

 

=  C 2 20 C 2 30 + C 2 22 C 2 30 - C 2 15 C 2 30 = 316 435