Solutions to Practice Problems for  the Lander Math Contest
Module 4, Solutions (10-12)

10) 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)=.7*.3^9 .
Similarly, we calculate probability that we are getting exactly
one head on the second toss:
P(T,H,8T)= (.3)(.7)(.3)^8 =.7*.3^9. We similarly get .7*.3^9 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/(.7*.3^9).
     b) Let us first calculate the probability of getting either no
heads or 1 heads. The probability of getting no heads is .
3^10. The probability of getting 1 heads is (as we showed
above) 10/(.7*.3^9). Thus the probability of getting either 0 or
1 heads is
.3^10+ 10/(.7*.3^9)
Since the probability of getting either 0 or 1 heads or at least
2 heads = 1, we get the equation .3^10+ 10/(.7*.3^9) + P(at
least 2 heads) =1, or transposing
P(at least 2 heads) = 1 - .3^10 - 10/(.7*.3^9)

11) a)The probability of such event is one specific choice
over the total number of possibilities. The total number would
be 14 for president, times 13 for vice president, times 12 for
treasurer, times 11 for the secretary:
P(Esther-P, Chana-VP, Rivka-T, Sarah-S)=1/(14*13*12*11)
     b) The total number of arrangements of 4 people is 4!=12,
the probability is
P(Esther, Chana, Rivka, Sarah to be P,VP,T,S)

12) The probability of 4 kings in the hand of 12 can be found
as a ratio of all hands with 4 kings (and 8 other cards) over
the total number of hands of 12.
The total number of hands of 12 would be found in the
following manner: the first card is picked up out of 36, times
the second card picked up out of 35, and so on: 36*35*…*25.
This would be the total number of hands if order matters, but
in our example order does not matter, so we have to divide
the previous result by 12! – the total number of possible
arrangements of 12 cards. In a hand with 4 kings we have
chosen 4 cards, the remaining 8 positions would be occupied
in the following ways: 32*31*30*29*28*27*26*25/8!-(it is
divided by 8! because order does not matter).
After canceling out the same numbers in the numerator and
denominator, our probability of getting 4 kings in a hand of 12
will be:
P(4K from 12)= 12*11*10*9/(36*35*34*33)=1/(7*17)=1/119.