Solution to homework #4, ICS6A, Winter, 1999
-- By Zhao, Hong

Textbook:

P291:
26. How many different strings can be made from the letters in MISSISSIPPI, using all the letters?
Answer: Since there are 1 "M", 4 "I"s, 4 "S"s, 2 "P"s -- totally 11 letters, according to THEOREM 3 on page 288, we can get the final answer:

11! / ( 1! * 4! * 4! * 2! ) = 34650
 

33. A student has three mangos, two papayas, and two kiwi fruits. If the student eats one piece of fruit each day, and only the type of fruit matters, in how many different ways can these fruits be consumed?
Answer: Similar to problem 26. There are total 7 pieces of fruit. The final answer:

7! / ( 3! * 2! * 2! ) = 210
 

P266:
10. What is the probability that a five-card poker hand contains the two of diamonds and the three of spades?
Answer: The total number of ways of the hand of cards without any constraints is: C( 52, 5 ).
The number of ways of the hand of cards which have to include 2 of diamonds and 3 of spades is:

C( 52-2, 5-2 ) = C( 50, 3 ).
So the probability is:
C(50, 3)/C(52, 5) = 0.00754
 

22. What is the probability that a positive integer not exceeding 100 selected at random is divisible by 3?
Answer: There are totally 100 positive integers not exceeding 100, which are: 1, 2, 3, ..., 100.
There are 33 positive integers that are not exceeding 100 and are divisible by 3, which are: 3, 6, 9, ..., 99. So the probability is: 33/100 = 0.33
 

33. Which is more likely, rolling a total of 9 when two dice are rolled or rolling a total of 9 when three  dice are rolled?
Answer: 9 = 3+6 = 4+5 = 5+4 = 6+3, -- 4 ways to get total of 9 points rolling two dice, so the probability that the outcome is 9 points is: 4/(6*6) = 1/9 = 0.111.
On the other hand:

So totally there are: 5+6+5+4+3+2 = 25 ways to get total of 9 points rolling three dice, so the probability that the outcome is 9 points is: 25/(6*6*6) = 0.116 > 0.111
So rolling 2 dice is more likely to get a total of 9 points.
 

P280:
3. Find the probability of each outcome when a biased die is rolled, if rolling a 2 or rolling a 4 is three times as likely as rolling each of the other four numbers on the die and it is equally likely to roll a 2 or a 4.
Answer: Let P(1) = P(3) = P(5) = P(6) = x, then P(2) = P(4) = 3x.
Because: P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = 1,
so: x+2x+x+2x+x+x = 10x = 1,
such that x = 1/10 = 0.1
Finally: P(1) = P(3) = P(5) = P(6) = 0.1 and P(2) = P(4) = 0.3
 

6. Suppose that E and F are events such that p(E) = 0.8 and p(F) = 0.6. Show that:

p(E and F) >= 0.4.
Answer: From THEOREM 2 on page 265, we have:
p(E or F) = p(E) + p(F) - p(E and F).
Then we have: p(E and F) = p(E) + p(F) - p(E or F)
For any probability, it will never exceed 1, so p(E or F) <= 1,
such that: p(E and F) >= p(E) + p(F) - 1 = 0.8+0.6-1 = 0.4
Finally, we have p(E and F) >= 0.4.
 

15. What is the conditional probability that a randomly generated bit string of length four contains at least two consecutive 0s, given that the first bit is  a 1? ( Assume the probabilities of a 0 and a 1 are the same. )
Answer: Let's name the bit string beginning with "1" is as event "A", and the bit string with at least 2 consecutive 0s as "B". So we can write:

 Thus, P(B|A) = P(A and B)/P(A) = 3/8
 

18. Assume that the probability a child is a boy is 0.51 and that the sexes of children born into a family are independent. What is the probability that a family of five children has

    a) exactly three boys?
    b) at least one boy?
    c) at least one girl?
    d) all children of the same sex?
Answer: According to THEOREM 1 on page 274:
a) P(a) = C(5, 3) * (0.51)3 * (1-0.51)5-3 = 0.318
b) P(b) = 1 - P(All girls) = 1 - (1-0.51)5 = 0.972
c) P(c) = 1 - P(All boys) = 1 - 0.515 = 0.965
d) P(d) = P(All girls) + P(All boys) = (1-0.51)5 + 0.515 = 0.063
28. A coin is biased so that the probability a head comes up when it is flipped is 0.6. What is the expected number of heads that come up when it is flipped 10 times?
Answer: The 10 trials are independent, and for each trial the expected value of number of heads is the same, which is E(n) = 1*0.6 + 0*(0.4) = 0.6. So we have:

32. What is the expected sum of the numbers that appear when three fair dice are rolled?
Answer: For a single die, the probabilities that the out come is 1 or 2, or3, ..., or 6 are the same, which is 1/6. So the expected value of the outcome is:

E(n) = 1*1/6 + 2*1/6 + 3*1/6 + 4*1/6 + 5*1/6 + 6*1/6 = (1+2+3+4+5+6)/6 = 3.5
For there independent fair dice the expected value of the sum of the outcomes is:
E(n1+n2+n3) = E(n1) + E(n2) + E(n3) = 3.5 + 3.5 +3.5 = 10.5