CBSE [Delhi]_XII_Mathematics_2004_Set I

Total number of balls = 7(white) + 5(black) + 3(red) = 15

Two balls are drawn out of 15 balls in ¹⁵C₂ ways.
Let us define the events as follows:

(i) A = Both the balls are red

There are 3 red balls out of which 2 red balls can be drawn in ³C₂ ways.

Thus, P(A) = ³C₂ /¹⁵C₂ =1/35.

(ii) B = Two balls are drawn, out of them one ball is red and the other is black

There are 3 red balls and 5 black balls out of which one ball drawn is red and the other ball drawn is black in ³C₁  x ⁵C₁ways.

Thus, P(B) =  ³C₁  x ⁵C₁/ ¹⁵C₂ = 1/7.

(iii) C = Two balls are drawn, out of them one ball is white.

There are 7 white balls and 8 non-white balls. Two balls are drawn out of them one ball is white and the other ball is non-white in  ⁷C₁  x ⁸C₁ ways.

Thus, P(C) =  ⁷C₁  x ⁸C₁/ ¹⁵C₂= 8/15.

Let X  denote the number of successes when a fair die is tossed twice. Then X is a random variable that can take the values 0, 1 or 2.

Probability that a number appearing on the top is less than 3 in a single throw of a dice = Probability of a success = 2/6= 1/3.

Probability of not getting a success = 1-1/3= 2/3

P( X = 0) = P( no success) = (2/3)(2/3)= 4/9

P( X = 1) = P(one success ) = (1/3) (2/3)+(1/3)(2/3) = 4/9

P( X = 2) = P( two success) =(1/3)(1/3)= 1/9

Given y² = a(b -x²)                                              ...(1)

Differentiating with respect to x, we get