prob. that a number greater than 4 appears = 2/6 = 1/3

prob. that a number less than or equal to 4 appears = 2/3



prob. that an even no. of tosses are needed =

(2/3)*1/3 + (2/3)^3*1/3 + (2/3)^5 * 1/3 ......



= 2/9 (1 + (2/3)^2 + (2/3)^4 + (2/3)^6 + ....)

= 2/9 [ 1 / ( 5/9) ]

= (2/9)*(9/5)

= 2/5

 (2) P(x)=a₀+a₁x+a₂x²............a_nx ⁿ, a_n # 0 be such that P(x²)=(P(x))^2. than value of P(x) =

P(1²) = [ P(1) ]²

Let n = 1

a0 = a0²

a0 = 1

LEt n = 2

a0  + a1 = a0² + a1² + 2.a0.a1

1 + a1 = 1 + a1² + 2a1

-a1 = a1²

a1 = 0, -1

as an # 0 ... so only solution is a1 = -1

so P(1) for n = 2 is 1 - 1 = 0

Similarly by solving ahead, we observe that:

P(x) = 0

 But look at this 

:\ 

:?

.