   Solutions to Practice Problems for  the Lander Math Contest
Module 3, Solutions (7-9)

7) The probability of drawing a king is 4/52=1/13.
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
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:             