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 

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

Let S be the sample space.

Then, n(S) = 52 C 2=(52 x 51)/(2 x 1) = 1326.

Let E = event of getting 1 spade and 1 heart.

n(E)= number of ways of choosing 1 spade out of 13 and 1 heart out of 13 =  13 C 1 * 13 C 1 = 169.

P(E) = n(E)/n(S) = 169/1326 = 13/102.

Let number of balls = (6 + 8) = 14.

Number of white balls = 8.

P (drawing a white ball) = 8 /14 = 4/7.

Here S= {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}.

Let E = event of getting at least two heads = {THH, HTH, HHT, HHH}.

P(E) = n(E) / n(S)

      = 4/8= 1/2

Clearly, there are 52 cards, out of which there are 12 face cards.

P (getting a face card) = 12/52=3/13.

P( selecting atleast one couple) = 1 - P(selecting none of the couples for the prize) 

 =  1 - 16 C 1 × 14 C 1 × 12 C 1 × 10 C 1 16 C 4 = 15 39

P(first letter is not vowel) =  2 4

P(second letter is not vowel) =  1 3

So, probability that none of letters would be vowels is =  2 4 × 1 3 = 1 6