# Maths-1999-Set I

(i) All Questions are compulsory

(ii) Question number 1 to 15 are of 2 marks each

(iii) Question number 16 to 25 are of 4 marks each

(iv) Question number 26 to 30 are of 6 marks each

**Q1) A coin is tossed 12 times. What is the Probability of getting exactly 8 tails? (Marks 2)**

Ans1) When a coin is tossed, we have S = {H, T}

p = P (getting a tail) = 1/2

… q = 1 – p = 1/2

Here n = 12

Now the Probability Distribution

B(n, p) = B(12, 1/2)

i.e. (q + p)12 where q = p = 1/2

… P (X = = 12C8 . q12 – 8 . p8

= 12C4 . q4 p8 = 12×11x10×9 x (1/2)4 + 8

4

= 498 = 495

212 4096

**Q2) Three coins are tossed simultaneously. List the sample space for the event. (Marks 2)**

Ans2) Sample space S, of tossing 3 coins simultaneously is given by:-

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

**Q3) Two cards are drawn without replacement from a well shuffled pack of 52 cards. What is the probability that one is red queen and the other is a king of black colour? (Marks 2)**

Ans3) We know that there are 2 red queens and two kings of black colour in a pack of 52 cards.

P (drawing a red queen) = 2/52 = 1/26

Since the card is not replaced, after the firs card, we have 51 cards left.

… P (drawing a black king) = 2/51

But these two draws are interchangeable.

… The reqd. probability = 2 x 1/26 x 2/51 = 2/(13 x 51) = 2/663

**Q4) Find a unit vector perpendicular to both = 3 + – 2 and = 2 + 3 – (Marks 2)**

Ans4) Here, = 3 + – 2 and = 2 + 3 -

=> x =

3 1 -2

2 3 -1

= 5 – + 7

… = Unit vector | to and

= ( x )/[ x ] = (5 – + 7)/(52 + (-1)2 + 72)

= (5 – + 7)/(75) = 1/(53) (5 – + 7)