Typical Test Problems (with solutions)

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Typical Test Problems (with solutions)
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Try solving these problems and compare with the
solutions
1. One bag contains 4 white balls and 2 black
balls another contains 3 white and 5 black. If
one ball is drawn from each bag, find the
probability that (a) both are white (b) both
are black (c) one is white and one is black.
Answer (a) 4/63/8 1/4 (b) 2/65/85/24 (c)
(4/6)(5/8)(2/6)(3/8) 13/24
2. A and B play a game in which they alternately
toss a pair of dice. The one who is first to get
a total of 7 wins the game. Find the probability
that (a) one who tosses first will win the game
(b) one who tosses second will win the game.
Reminder Infinite Geometric Progression
Answer. (a) (1/6)(5/6)²(1/6)(5/6)4(1/6)...6/
11 (b) (5/6)(1/6) (5/6)³(1/6)....5/11
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3. A machine produces a total of 12,000
bolts a day, which are on average 3 defective.
Find the probability that out of 600 bolts
chosen at random, 12 will be defective.
Solution, Of 12000 bolts, 360 are defective and
11640 are not. P C11640,588 C360,12
/C12000,600 0.0347277
Some comments This problem caused some
controversy. Some of you suggested that the
probability of finding a defective bolt is 0.03.
Then the probability could be estimated through
the Binomial distribution p(12 out of 600)
C600,12 0.03120.975880.036 (Eq. 2). This is
very close indeed to the result of Eq. 1. May be
this tiny difference is just a result of an
error? To demonstrate that this is not true, we
consider a simpler example. Suppose that there
are 10 balls, 6 are black and 4 are white. What
is the probability that of 4 balls chosen at
random, 2 are white.
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The conventional solution is p
C4,2C6,2/C6,40.429. Trying to be silly, we can
also use a Binomial formula assuming that the
probabilities are 3/5 and 2/5 p
C4,2(2/5)2(3/5)2 0.346. This time the
difference is significant. In addition, we can
now understand the source of the problem, The
Binomial distribution can be used only when the
probabilities of two outcomes do not depend on
the number of previous trials. This condition
is not fulfilled in our example. The probability
of choosing a white ball depends on the outcomes
of previous selections. The same is true for the
bolts problem, although the differences are
much less significant. Suggestion try to
investigate how the results derived with Eqs. 1
and 2 depend on the total number of bolts n
(initially we had n12000). Change n from 100 to
100000. Use Matematica. The function below
utilizes Eq. 1 and can be useful. We still
assume that 600 bolts are selected, and the
probability of 12 defective bolts is
questioned. fn_ Binomial0.03n,12Binomial0
.97 n, 588/Binomialn,600//N Please, let me
know about the outcome. Good luck.
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4. The probabilities that a husband and wife will
be alive 30 years from now are given by 0.7 and
0.8 respectively. Find the probability that in 20
years (a) both (b) neither (c) at least one, will
be alive.
Answer (a) 0.56 (b) 0.06 (c) 0.94
5. Four different mathematics books, six
different physics books, and two different
chemistry books are to be arranged on a shelf.
How many different arrangements are possible if
(a) the books in each particular subject must all
stand together. (b) only the mathematics books
must stand together
Answer (a) 6!4!23! 207360 (b) 12-41 9
9!4! 8709120
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6. An urn contains 6 red and 8 blue marbles. Five
marbles are drawn at random from it without
replacement. Find the probability that 3 are red
and 2 are blue.

• Answer C6,3 C8,2/C14,50.27972
• Some comments This problem caused a very
  productive discussion in one of my classes. The
  counterargument to our solution sounded as
  followsLets consider one sequence of choices
  leading to 3 r and 2 b balls. Its probability
  is
• P(r,r,r,b,b) (6/14)(5/13)(4/12)(8/11)(7/10). For
  any other sequence (such as r-b-r-b-r) the
  probability is the same. For instance,
• P(r,b,r,b,r) (6/14)(8/13)(5/12)(7/11)(4/10)
  P(r,r,r,b,b) 0.027972 . Based on this, it was
  suggested that this probability (which turned to
  be 1/10 of the previous result), gives the
  correct answer to our problem.
• This error has a historical parallel , the
  famous D Alamberts assertion. DAlambert did
  not notice that the Head and Tail event for 2
  coins corresponds to two outcomes HT and TH.
  However, DAlambert wrongly suggested that the
  probability of H and T is 1/3, while our
  students calculated the probabilities of each
  sequence correctly. They simply did not sum up
  the probabilities of various outcomes
  corresponding to the event in question (3 red and
  2 blue).
• In our case, the number of outcomes is the
  number of possible arrangements of 3 r and 2 b
  marbles C5,3C5,210. Thus, the total
  probability is
• p(total) 10 P(r,r,r,b,b) 0.27972.

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7 (a) Find the probability of getting the sum 7
on at least one of three tosses of a pair of fair
dice. (b) How many tosses (n) is needed in
order that this probability (getting 7 in at
least 1of n tosses) is greater than 0.95.
Answer (a) 1-(5/6)³91/216(b)
(5/6)nlt0.05,n17.
8. The odds in favor of A winning a game of chess
against B are 32. If three games are to be
played, what are the odds (a) in favor of A
winning at least 2 games (b) against A of losing
the first two games to B
Answer (a) 3(3/5)²(2/5)(3/5)³ 81/125
(8144)(b) 4/25 (421)
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9. Out of 800 families with 4 children each, how
many would be expected to (a) 2 boys and 2 girls,
(b) at least 1 boy (c) no girls (d) at most 2
girls
Answer the probabilities are (a)C4,20.54
3/8. 0.375 (b)1-(1/2.)4 0.9375 (c)(1/2.)4
0.0625 (d) F BBBB BBBG BBGG P
0.37540.0625 0.0625 0.6875. The rest is
obvious.
10. A pair of dice is tossed repeatedly. (a) Find
the probability that the sum of 11 occurs for the
first time on the 6-th toss.
Answer (1/18)(17/18)5 0.262
11. Find the probability of scoring a total of 7
points (a) once, (b) at least once, (c) twice, in
two tosses of a pair of dice.
Answer (a) 2(1/6)(5/6) 5/18 (b) 1-25/36
11/36 (c) 1/36
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12. A box contains 9 tickets numbered from 1 to
9, inclusive. If 3 tickets are drawn from the box
1 at a time, find the probability that they are
alternatively either odd-even-odd or even-odd
even. Solution Total number of choices
P_9,39!/6!504 (Notice that the order here is
important).The number of the possible choices for
the first event 544, For the second, 453.
The answer (544453)/504 5/(18) 13. Three
cards are drawn from an ordinary deck of 52
cards. Find the probability that (a) all cards
are of the same suit (b) at least two aces are
drawn Answer (a) There are Binomial13, 3
(I use this Mathematicas name for Cn,m) ways of
choosing 3 cards from one suit (say hearts). It
should be multiplied by 4, the total number of
suits P 4.Binomial13, 3/Binomial52,
30.052 (b) There are Binomial4,248 ways
to chose 2 cards out of 4 aces and one card which
is not an ace, and Binomial4,2 ways to choose
three aces out of four P (Binomial4,248
Binomial4,3)/Binomial52, 30.013
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14. 15 students are evenly distributed
among three groups. If three of them are
siblings, (a) what is the probability that all
three are in the same group? (b) What is the
probability that each group gets one?
Solution (a) The total number of outcomes
N15!/(5!5!5!). The positive outcomes are 3
12!/(2!5!5!) (If you put all 3 in one of the
groups, than you have to distribute other 12 in
the groups of 5,5,and 2. And then it should be
multiplied by 3.) (b) 3! 12!/(4!4!4!)
15. In how many ways can 4 boys and 4 girls pair
off? In how many ways can they stay in a row in
alternating gender? Answer (a) 4! (b) 24!
4!
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16 A marble is drawn at random from a box
containing 10 red, 30 white, 20 blue and 15
orange marbles. Find a probability that it is
(a) orange or red (b) nor red or blue Answer
(a) 10/7515/75 1/3 (b) 45/75 3/5 17. Two
marbles are drawn at random with replacement from
a box containing 10 red, 30 white, 20 blue and 15
orange marbles. Find a probability that (a)
the first is blue and the second is orange (b)
they are either red or white or both (red and
white). Hint "both" consists of 2 outcomes,
depending on the order. Answer (a)
2015/(7575) 4/75 (b)
((10/75)²(30/75)²21030)/(7575) 64/225