\(m \leq 5\)
\(2 x^{2}+2 m x+m+1>0\)
For real roots
\(4 m^{2}-8(m+1)>0 \)
\(4 m^{2} - 8 m-8>0 \)
\(m^{2}-2 m-2>0 \)
\(\text {Then } m \text { can be } 5,4,3\)
\(n(S)=5 ,n(A)=3 ,P(A)=\frac{3}{5} \)

\(P(A)=0.3, \quad P(B)=0.6, \quad P(A \cap B)=0.18 \)
\(P(A \cup B)=P(A)+P(B)-P(A \cap B) \)
\(=0.3+0.6-0.18 =0.72 \)
\(P(\bar{A} \cup \bar{B})=1-P(A \cap B) \)
\(=1-0.72=0.28 \)

Probabilrty of getting atleast 8 heads
\(=P(8)+P(9)+P(10) \)
\(=10 C_{8}\left(\frac{1}{2}\right)^{10}+10 C_{9}\left(\frac{1}{2}\right)^{10}+10 C_{10}\left(\frac{1}{2}\right)^{10} \)
\(=\frac{45+10+1}{1024}=\frac{56}{1024}=\frac{7}{128} \)

\(P\left(A_{1}\right)=0.6, \quad P\left(A_{2}\right)=0.4\)
\(A_{1}=\text { Event of selecting a girl }\)
\(\mathrm{B}=\text { Event of student taller than } 1.8 \mathrm{~m}\)
\(A_{2}=\text { Event of selecting a boy }\)
\(P\left(B / A_{1}\right)=\frac{1}{100}, P\left(B / A_{2}\right)=\frac{4}{100}\)
To find \(=\frac{P\left(A_{1}\right) \cdot P\left(B / A_{1}\right)}{P\left(A_{1}\right) \cdot P\left(B / A_{1}\right)+P\left(A_{2}\right) \cdot P\left(B / A_{2}\right)} \)
\(=\frac{0.6 \times \frac{1}{100}}{0.6 \times \frac{1}{100}+0.4 \times \frac{4}{100}} \)
\(=\frac{6}{\frac{6}{1000}+\frac{16}{1000}} \)
\(=\frac{6}{22}=\frac{3}{11} \)

\(n(S)=n(A \times A) \quad=16\)
\(\text {To find } n(A)\)
\(\text {Let } a=4, b=1 \text { then } a^{2}-4 b \text { is }\)
A = 1,b = 1 then 16 - 4 > 0
b = 2. then 16 - 8 > 0
b=3 then 16-L2>0
b = 4 then 16 - 16 = 0
Let Q = 3,b = 1 thg,l, 9-4 > 0
b = 2 then 9 - 8 > 0
Let a = 2,b = l then 4 - 4 > 0
n(A).= 7,
\(P(A)=\frac{n(A)}{n(S)}=\frac{7}{16}\)

\(\mathrm{P}(\mathrm{B} / \mathrm{A}) =\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{A})} \Rightarrow 0.6=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{0.4} \)
\(\mathrm{P}(\mathrm{A} \cap \mathrm{B}) =0.6 \times 0.4 =0.24 \)
\(\mathrm{P}(\overline{\mathrm{A}} \cap \mathrm{B}) =\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \)
\(=0.8-0.24 =0.56 \)

\(\mathrm{P}(\mathrm{A} \cup \mathrm{B}) =\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \)
\(0.6 =0.35+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B}) \)
\(0.25 =\mathrm{P}(\mathrm{B})-0.35 \mathrm{P}(\mathrm{B}) \)
\(=\mathrm{P}(\mathrm{B})(1-0.35) \)
\(=\mathrm{P}(\mathrm{B})(0.65) \)
\(\mathrm{P}(\mathrm{B}) =\frac{0.25}{0.65}=\frac{5}{13} \)

The equation is not valid for 1 to 9. 30 to 49
(i.e) for 9 + 20 = 29 numbers
Valid for 10 to 29 and 50 to 100 = 71 numbers
n(A) = 71, n(S) = 100
\(\mathrm{P}(\mathrm{A})=\frac{71}{100}=0.71\)