a) If we replace the card, then the probability of

second drawing the king is the same. Finally, the

probability of the two kings in a row is

(1/13)(1/13)=1/169, which is approximately .0059 or

about .6%.

b) If we do not replace the first card, the second

drawing of the king will be 3/51=1/17, then the final

probability of two consequent kings is

(1/13)(1/17)=1/221.

8) a) We first find the possibility of a red marble:

P(R)=6/(4+3+6)= 6/13,

then the probability of the white after the red one:

P(B|R)=4/(4+3+5)=4/12=1/3,

and the black marble just after the red and the white:

P(W|R,B)=3/(3+3+5)=3/11. So, the resulting

probability of drawing red, white and black in this

order is the product of previous three probabilities:

P(R then B then W)=(6/13)(1/3)(3/11)=6/143.

b) Similar approach allows us to calculate the

probability of drawing red, white and again red:

P(R then W then Red)=(6/13)(3/12)(5/11)=15/143.

9) a) Consecutive tosses of a fair coin are

independent events. The probability of first one head

and then nine tails is equal to:

P(H,9T)=(1/2)(1/2)^9 =1/(2)^10.

Similarly, we calculate probability that we are getting

exactly one head on the second toss:

P(T,H,8T)= (1/2)( (1/2)(1/2)^10 =1/(2)^10.

We similarly get 1/(2)10 for just one head in each of

the other 8 possible positions.

To find the probability of the exactly one head (in all

possible arrangements with nine tails) we have to

add up these ten equal probabilities:

P(exactly one head)=10/(2)^10.

b) Let us first calculate the probability of getting

either no heads or 1 heads. The probability of getting

no heads is 1/(2)^10. The probability of getting 1

heads is (as we showed above) 10/(2)^10. Thus the

probability of getting either 0 or 1 heads is 1/(2)^10 +

10/(2)^10 = 11/(2)^10 . Since the probability of

getting either 0 or 1 heads or at least 2 heads = 1,

we get the equation

11/(2)^10 + P(at least 2 heads) =1, or transposing

P(at least 2 heads) = 1- 11/(2)^10 = 1013/1024.