1) Show it is more likely to throw one ace with four dice than to
throw at least one double ace in 24 throws of two dice.
2) Letters in the morse code are formed by a succession of dashes and
dots with repetions permitted. How many letters is it possible to form
with ten symbols or less?
3) The numbers 1,2,....,n are arranged in random order. Find the
probability that the symbols 1 and 2 appear as neighbors in the order
named. Do the same for the symbols 1,2 and 3.
4) From a population of five symbols a,b,c,d,e a sample of 25 is
taken. Find the probability that the sample will contain five symbols
of each kind.
I need answer for those four probability questions. Manhattan Review InFocus Newsletter::
You might see a few probability questions and one or two combinatorics questions. There were a few questions in which many of the incorrect answer choices were
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Dear al3in,
Let me address your questions in order. Bear in mind that my solutions
are intended for instructional purposes and that you are not to
represent them as your own work.
1.
The probability of throwing at least one ace with four dice is
complementary to the probability of throwing no aces with four dice.
With one die, the probability of not throwing an ace is
5/6 .
With four dice, the probability of throwing an ace with none of them is
(5/6)^4
so the probability of throwing at least one ace is
1 - (5/6)^4
or about
0.52 .
Similarly, the probability of throwing at least one double ace in 24
throws of two dice is complementary to the probability of throwing no
double aces in those 24 throws. The probability of not throwing a
double ace with two dice is
35/36
so the probability of not throwing a double ace in 24 throws of two dice is
(35/36)^24 .
The probability of throwing at least one double ace in 24 throws of
two dice is complementary to that:
1 - (35/36)^24
or about
0.49 .
Since 0.52 is greater than 0.49, it is more likely to throw one ace
with four dice than to throw at least one double ace in 24 throws of
two dice.
2.
Since the Morse alphabet contains two kinds of symbol, there are 2^k
ways to form a Morse code using exactly k symbols.
A Morse code of 0 symbols can be formed in
2^0 = 1
way, a Morse code of 1 symbol can be formed in
2^1 = 2
ways, one with 2 symbols in
2^2 = 4
ways, and so on. The remaining numbers of ways to form a Morse code
with exactly k symbols, where k is no greater than 10, are shown
below.
2^3 = 8
2^4 = 16
2^5 = 32
2^6 = 64
2^7 = 128
2^8 = 256
2^9 = 512
2^10 = 1024
The sum of all these amounts to the total number of ways there are to
form a Morse code with 10 symbols or fewer. As an alternative to
summing them directly, we can take advantage of the fact that the sum
of all powers of 2 with exponent less than n is equal to
2^n - 1 .
Thus, the sum of all powers of 2 with exponent less than 11 is
2^11 - 1
or
2048 - 1 = 2047 .
In conclusion, there are 2047 Morse codes made of ten symbols or fewer.
3.
There are 2! = 2 ways for the symbols 1 and 2 to neighbor each other.
Furthermore, a subsequence of two symbols can appear in n-1 places
within a sequence of n symbols. In each of these cases, the remaining
n-2 symbols of the sequence can be ordered in (n-2)! ways. Thus, the
number of ways for symbols 1 and 2 to neighbor each other within a
sequence of n symbols is
2! * (n-1) * (n-2)! .
In total, there are n! ways to order n symbols. The probability that
symbols 1 and 2 will neighbor each other within such a sequence is
therefore
2! * (n-1) * (n-2)! 2 * (n-1)! 2
------------------- = ---------- = -
n! n * (n-1)! n
or simply
2/n .
Similarly, there are 3! ways for the symbols 1, 2, 3 to appear as
neighbors, and such a subsequence can occur in n-2 positions within a
sequence of n symbols. There are (n-3)! ways for the remaining symbols
to be ordered. Hence, the probability of symbols 1, 2, 3 neighboring
one another in a random ordering of n symbols is
3! * (n-2) * (n-3)! 6 * (n-2)! 6
------------------- = ------------------ = ---------
n! n * (n-1) * (n-2)! n * (n-1)
or simply
6/(n^2-n) .
4.
Recall that the number of ways to choose m items from a set of n is written
C(n, m), where
n!
C(n, m) = ----------- .
(n-m)! * m!
We are now faced with the task of enumerating the ways in which a
sequence of 25 symbols, drawn from a population of 5 types of symbol,
contains 5 symbols of each kind. The symbols are denoted a, b, c, d,
e.
Without loss of generality, we can begin by finding positions in the
sequence for the 5 symbols a. There are C(25, 5) choices of 5
positions among the 25 total. Next, let us find positions for the 5
symbols b. There are C(20, 5) ways to find 5 positions among the
remaining 20. Similarly, there are C(15, 5) ways to distribute the c's
and C(10, 5) ways for the d's, whereafter the distribution of e's is
uniquely determined.
The total number of ways to draw 25 symbols from a population of 5
types of symbol is 5^25. Hence, assuming that the sample of 25 is
drawn from an evenly distributed population, the probability of
drawing 5 symbols of each type is
C(25, 5) * C(20, 5) * C(15, 5) * C(10, 5)
-----------------------------------------
5^25
or about
0.0021 .
And that's the lot!
If you feel that my answer is incomplete or inaccurate in any way, please
post a clarification request so that I have a chance to meet your needs
before you assign a rating.
Regards,
leapinglizard A couple questions about Probability? I have a few questions that I am ::
1. Five The phrase is misleading! It does not say how many outcomes are possible in total? it into distinct probabilities: the probability that out of 28
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NOAA Technical Memorandum NWS AR-44::
from this pair of questions were compared to look for Numerous received an average probability of 72.3%, and areas of only received a 43.1% mean POP.
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Probability- A few questions