A bag contains 8 balls (black and white). If four balls are chosen without replacement then 2W and 2B are found then the probability that number of white and black balls are same in bag is equal to

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  $=\frac{30}{11*11}=\frac{30}{121}$

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 = $\frac{5}{216}$   

Let

A : Missile hit the target

B : Missile intercepted

P (B) = $\frac{1}{3}P\left(A/\overline{B}\right)=\frac{3}{4}$ $\frac{}{}\stackrel{}{}\frac{}{}$

$P\left(\overline{B}\right)=\frac{2}{3}$

$\Rightarrow P\left(\overline{B}\cap A\right)=\frac{3}{4}\times \frac{2}{3}=\frac{1}{2}$

Required Probability = $\frac{2}{3}\times \frac{3}{4}\times \frac{2}{3}\times \frac{3}{4}\times \frac{2}{3}\times \frac{3}{4}=\frac{1}{8}$