90. If each element of a second order determinant is either zero or one, what is the probability that the value of the determinant is positive? (Assume that the individual entries of the determinant are chosen independently, each value being assumed with probability 1/2).

P (2 obtained on even numbered toss) = k (let)

P (2) = $\frac{1}{6}$  

P (  $\overline{2}$ )= $\frac{5}{6}$  

$k=\frac{5}{6}\times \frac{1}{6}+{\left(\frac{5}{6}\right)}^3\times \frac{1}{6}+{\left(\frac{5}{6}\right)}^5\times \frac{1}{6}+...$

$=\frac{\frac{5}{6}\times \frac{1}{6}}{1-{\left(\frac{5}{6}\right)}^2}$

$=\frac{5}{11}$

If x = 0, y = 6, 7, 8, 9, 10

If x = 1, y = 7, 8, 9, 10

If x = 2, y = 8, 9, 10

If x = 3, y = 9, 10

If x = 4, y = 10

If x = 5, y = no possible value

Total possible ways = (5 + 4 + 3 + 2 + 1) * 2

= 30

Required probability  $=\frac{30}{11*11}=\frac{30}{121}$

P (2W and 2B) = P (2B, 6W) × P (2W and 2B)

+ P (3B, 5W) × P (2W and 2B)

+ P (4B, 4W) × P (2W and 2B)

+ P (5B, 3W) × P (2W and 2B)

+ P (6B, 2W) × P (2W and 2B)

(15 + 30 + 36 + 30 + 15)

$=\frac{36}{126}$

$=\frac{18}{63}$

$=\frac{6}{21}$

$=\frac{2}{7}$

Let probability of tail is   $\frac{1}{3}$

⇒ Probability of getting head = $\frac{2}{3}$  

∴ Probability of getting 2 heads and 1 tail

$=\left(\frac{2}{3}\times \frac{2}{3}\times \frac{1}{3}\right)\times 3$

$=\frac{4}{27}\times 3$

$=\frac{4}{9}$                  

                  &nb

ax² + bx + c = 0

D = b² – 4ac

D = 0

b² – 4ac = 0

b² = 4ac

(i) AC = 1, b = 2 (1, 2, 1) is one way

(ii) AC = 4, b = 4

$\begin{array}{l}a=4c=1\\ a=2c=2\\ a=1c=4\end{array}\}3ways$

(iii) AC = 9, b = 6, a = 3, c = 3 is one way

1 + 3 + 1 = 5 way

Required probability = $\frac{5}{216}$