   Solutions to Practice Problems for  the Lander Math Contest
Module 2, Solutions (4-6)

4)         The total number of outcomes for two dice being
tossed is (6)(6)=36. The number of outcomes that the first die
shows a 3 and the second die shows NOT a 3 is five: (3,1),
(3,2), (3,4), (3,5) and (3,6). The same number of outcomes –
five, is for the second die to show the 3, and the first die to be
different from 3. Then, the probability of both dice to show
exactly one 3 is:
P(exactly one 3) =P(3 on 1st die, not 3 on 2nd die) + P(not
3 on 1st die, 3 on 2nd die =5/36+5/36=10/36.
There is the only one event to get exactly two 3s – (3,3); the
corresponding probability is P(two 3s) = 1/36.
The sought for probability is the sum of the probability of
showing exactly one 3 and the probability of showing exactly
two 3s’ (in two tosses):P(at least one 3) = P(exactly one 3) + P
(two 3) = 10/36 + 1/36 = 11/36

5)         Let us trace the possibility for the marble to be black.
The probability of black marble taken from the red bag is 4/7.
If the coin shows the heads (P(H)=1/2), then the probability of
both events is
P(H)P(Black marble from red bag)=(1/2)(4/7)=2/7
Consequently, the probability of black marble taken from the
yellow bag is 2/7. If the coin shows the tails (P(T)=1/2), then
the probability of both events is:
P(T)P(Black marble from yellow bag)=(1/2)(2/7)=1/7.
Finally, P(Black marble)= 2/7+1/7=3/7

6)         The probability of two heads followed by two tails (in
this special order) is P(HHTT)=(1/2)^2(1/2)^2 =1/16. There
are 6 possible arrangements of two heads and two tails:
HHTT, HTHT, HTTH, THHT, THTH and TTHH with the same
probability of 1/6. The probability of two heads and two tails
without specifying the order is: P=6(1/16)=3/8             