HW2

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HW2

• assignment

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• A fair coin is tossed 5 times. Find a probability
  of obtaining (a) exactly 3 heads (b) at most 3
  heads SolutionBinomial distribution.(a)
  P(n3) C5,3 0.530.52 Binomial5,30.55
  0.31(b) P(nlt3) 1-P(4)-P(5) 0.81
• A hat contains tags numbered 1,2,3,4 and 5. A tag
  is drawn from the hat and replaced, then the
  second tag is drawn. (a) show the sample
  space(b) Find the probability that the number
  on the second tag exceeds the number on the
  first.Solution (a) S includes 25 points,
  x,y, Similar to a sample space for two dice
  with 5 faces each (b) There are 10 points in S
  where ygtx. P(ygtx) 10/25 2/5.(c) Find the
  probability that the first tag has a prime number
  and the second tag has an even number (1 is not
  considered a prime number)Solution The points
  are 22,24,32,34,52,54. P6/25
  0.24.
• There are 4 roads connecting the cities A and B
  and 3 roads between the cities B and C. How many
  ways can round trip be made from city A to C
  (through B) and back? How many ways if it is
  desired to take different routs on the way back?
  Solution 4334 144, 4323 72.

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• (a) How many license plates are possible if the
  first three places are occupied by letters and
  the last three by numbers(b) Assuming all
  combinations are equally likely, what is the
  probability the three letters and the three
  numbers are different?Solution(a) 263 103
  (b) P26,3P10,3/ 263 103 0.64
• Six students, three boys and three girls, line up
  in a random order for a photograph. Whats the
  probability that the boys and girls
  alternate?SolutionN 3!3! 3!3! 72 P
  72/6! 72/720 0.1

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6. Ten children are to be grouped into two clubs
(Lions and Tigers) with 5 children in each club.
Each club is then to elect a president and a
secretary. In how many ways can this be done?
SolutionC10,5P5,2P5,2100800. 7. N
people are arranged in a line. What is the
probability that 2 particular people, say A and
B, are not next to each other?
Solution Consider two people as a unit. There
are (N-1)! Ways to permute such a group. The
total number of permutations corresponding to A
and B placed together is 2(N-1)!. The
probability P(A and B are together) 2(N-1)!/N!
2/N.The answer is P(A and B separated) 1
2/N (N-2)/N
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• How many ways can 4 couples sit at a round table
  if each man sits next to two women neither one of
  whom is his wifePlacing men in every other chair
  (even or odd) results in 2 4! possibilities.Thus,
  we have 48 possible arrangements. For each of
  these arrangement, for every empty chair there
  are two possible choices of a woman per such
  that she is married to none of her two neighbors.
  Solution(1) Lets place men first. The two
  possibilities are (a) every other odd chair (b)
  every other even chair. The total number of
  arrangements is 24! 48 (2) The second step
  is to place the women in the empty chairs. For
  each placement of the men, there are two possible
  arrangements of the women that satisfy the
  conditions. Consider an exampleM1 W3 M2 W4
  M3 W1 M4 W2M1 W4 M2 W1 M3 W2 M4 W3
• The answer is N482 96.

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9. How many times should you roll a die so that
the probability of at least one 6 is at least
0.9? SolutionP(m,n) is the probability of m
sixs in n rolls. P(mgt1,n) 1-P(m0,n)P(0,n)
(5/6)nThis probability must be less or equal
to 0.1 (so that the probability of the compliment
event, at least one 6, is gt 0.9. Thus we have
to find an integer n with which P(0,n) is the
number closest to 0.1 from below. We can first
solve the equation (5/6)n 0.1 n
Log0.1/Log5/6 12.6 The closest integer
solution is 13. You can also do it with
Mathematica by making a table of P(0,n) for
n,0,13
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10. A student has no knowledge whatsoever of the
material to be tested on a true-false
examination, and so the student flips the coin in
order to determine the response to each question.
What is the probability that the student scores
at least 60 on a ten-item examination? Solution
At least 6 means 6 OR 7 OR 8 OR 9 OR 10. The
events are disjoined and the probability equals
the sum of the individual probabilities 0.510
(Binomial10,6Binomila10,7
Binomial10,10)0.377
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11. If 70 of the bolts produced by a machine are
perfect, 20 are of average quality and 10
are defective, determine the probability that out
of 15 bolts chosen at random, (a) exactly two are
average, (b) 10 are perfect, 3 are average and 2
are defective, (c) less than 3 are defective.
11. Solution (a) Exactly two are average.In
this case the problem is reduced to Binomial
distribution with m2, p0.2,q0.8 P(2 average)
Binomial2,15 (0.2)2 (0.8)13 0.23 (b) 10
are perfect, 3 are average and 2 are
defective. Now we have three outcomes, and have
to apply a multinomial distributionP(10,3,2)
(15!/10!3!2!) 0.7100.230.12 0.07 (c) Less than
3 are defective. Binomial, with p 0.1 and q
0.9. P(ndeflt3) P(0)P(1)P(2)
Binomial15,00.915 Binomial15,10.9140.1
Binomial15,20.9130.12 0.82