Probability

(80+1)? /8 = (¹C?80? +.+¹C? )/8. Remainder is 1/8.

Required probability = (? C? +? C? )/¹¹C? = 25/165=5/33.

Required area = 4 [∫? ¹/² 2y²dy + ½*½*1] = 5/6.

Three balls can be given to B? in? C? ways. Now remaining 6 balls can be distributed into 3 boxes in 3? ways.
Total no. of favourable events =? C? * 3? = 28 * 3?
Total no. of events = 9 balls distributed into 4 boxes in 4? ways.
probability = 28 * 3? /4? = 28/9 * (3/4)? ⇒ k = 28/9
k ∈ |x-3|<1

Probability (at most two machines will be out of service) = (3/4)³ . k
⇒ ?C?(1/4)?(3/4)? + ?C?(1/4)¹(3/4)? + ?C?(1/4)²(3/4)³ = (3/4)³ . k
⇒ 17/8 (3/4)³ = (3/4)³ . k
⇒ k = 17/8

A and B are independent events so P(A/B) = 1/3

P (at least one head) = 1 - P (no heads) = 1 - (1/2)? ≥ 0.9.
0.1 ≥ (1/2)?
10 ≤ 2?
n=3, 2³=8. n=4, 2? =16.
Minimum value of n is 4.

(2? -2) is multiple of 3.
If n is odd, 2? ≡ (-1)? =-1 (mod 3). 2? -2 ≡ -1-2 = -3 ≡ 0 (mod 3).
If n is even, 2? ≡ (-1)? =1 (mod 3). 2? -2 ≡ 1-2 = -1 (mod 3).
So n must be odd.
2-digit numbers are 10-99 (90 numbers).
Odd numbers are 11,13, .,99. Number of terms = (99-11)/2 + 1 = 45.
Probability = 45/90 = 1/2.

Circle: (x-1)² + (y-3)² = 4. C= (1,3), r=2.
Length of tangent from P (-1,1) is L = √ (-1)²+1²-2 (-1)-6 (1)+6) = √ (1+1+2-6+6) = √4 = 2.

Area of quadrilateral PACB = 2 * Area (PAC) = 2 * (1/2 * L * r) = 2*2=4.
AB is chord of contact. T=0 => -x+y- (x-1)-3 (y-1)+6=0 => -2x-2y+10=0 => x+y=5.
Distance of C from AB = |1+3-5|/√2 = 1/√2.
Length of AB = 2√ (r²-d²) = 2√ (4-1/2) = 2√ (7/2) = √14.
Area of ABD =?

Mean = Σx? p? = 0 (1/2) + Σ? ∞ j (1/3)? = (1/3)/ (1-1/3)² = (1/3)/ (4/9) = 3/4
P (X is positive and even) = P (X=2) + P (X=4) + .
= (1/3)² + (1/3)? + . = (1/9)/ (1-1/9) = (1/9)/ (8/9) = 1/8