If an unbiased coin is tossed four times, are the following the events a and b conditionally independent given c : a={there is at least one tail in the first 3 tosses} b ={there is at least one head in the last 3 tosses} c = {the second and third toss respectively reveal heads, tails}

A and C are not independent.

B and C are not independent either.

Step-by-step explanation:

By the multiplication rule, the sampling space has possible outcomes.  

Let's compute P(A), P(B), P(C), P(A∩C), P(B∩C), P(A|C) and P(B|C).

As there are only two outcomes that does not have at least on tail in the first three tosses (H,H,H,H) and (H,H,H,T) then  

P(A) = 14/16 = 7/8

Similarly, as there are only two outcomes that does not have at least on head in the last three tosses (H,T,T,T) and (T,T,T,T) then  

P(A) = 14/16 = 7/8

A∩C and B∩C have 4 possible outcomes (T,H,T,T), (T,H,T,H), (H,H,T,T) and (H,H,T,H), so  

P(A∩C) = P(B∩C) = 4/16 = ¼

P(A) given C and P(B) given C have the following probabilities:  

Since P(A|C)≠P(A), A and C are not independent

Since P(B|C)≠P(B), B and C are not independent

then, the surface area of the sphere shown in the image is: 256π sq. units!

step-by-step explanation:

the lateral surface area of a sphere is 2πrh where h=2r.

then:

surface area = 4πr^2.

in this case, r=8 units ⇒ surface area = 4π(8 units)^2. = 256π sq. units!

then, the surface area of the sphere shown in the image is: 256π sq. units!