Download Test #3 - Yeah, math, whatever.

Practice Test #3 MAT 120
(1) Experiment: Draw a card from a deck of 52. Find the probability that the card is:
(a) A black card.
(b) A diamond
(c) An Ace
(d) An Ace or a black card.
(e) A face card (King, Queen, Jack) or a club.
(2) Experiment: roll a pair of dice. Find the probability that you roll:
(a) 7 or 11
(b) Greater than 9
(c) 7, given that the roll is odd.
(3) An urn contains 5 blue balls and 4 red ones.
(a) If a ball is drawn at random, not replaced, and then another is drawn, what is the
probability that both balls were blue?
(b) If 3 balls are drawn without replacement, what is the probability that the first one is
blue, the second and third ones are red?
(4) If you draw three cards (in succession) from a deck, without replacement, what is the
probability that you will have a King, an Ace, and a King in that order?
(5) 22 percent of Americans don't have a clue about anything. If you ask 3 random
Americans about something, what is the probability that none of them have a clue? What
is the probability at least one of them has a clue?
(6) A bored Sveta decides to walk from point A to point C, stopping by B on the way.
She figures there are 4 routes from A to B, and 3 from B to C. How many combined
routes are there from A to C?
(7) K has 8 books, but only 5 will fit on a small shelf that he has. In how many ways can
he select and arrange the 5 books?
(8) An ID code is to be formed using 3 letters and 4 digits (numbers). In how many ways
can the code be formed if:
(a) Any letter or digit can be repeated?
(b) No letter or digit may be repeated?
(9) In how many ways can you deal 5 cards from a deck of 52?
(10) A Senate committee consists of 14 Senators, 8 of whom are Democrats, and 6
Republicans.
(a) A sub-committee is to be formed, consisting of 3 Democrats and 2 Republicans. In
how many ways can the committee be formed?
(b) Another sub-committee will have 3 Senators (parties not applicable). In how many
ways can this one be formed is one member will be chairman, one vice-chairman, and
one misinformation specialist?
(11) A five-card hand is dealt from a deck of 52. What is the probability that the hand:
(a) Is a diamond flush? (all 5 cards are diamonds)
(b) Has 3 Kings?
(c) Has 3 Aces, and 2 Kings?
(12) 8 men and 6 women form a staff in a corporate office. If 4 of them are indicted for
security fraud, what is the probability that:
(a) 2 are men, 2 are women?
(b) All four are men?
(c) They're all guilty? (that's a joke)
Solutions:
(1) (a) There are 26 black cards in the deck (the spades and the clubs). Thus the
26
 .5
probability is P(Eblack) =
52
(b) There are 13 diamonds in the deck, so P(E♦) =
(c) 4 aces, so P(Eace) =
13
 .25
52
4
1

52 13
(d) We have to add the probabilities of an ace or a black card, then subtract the
probability of the 'overlap' (the two black aces)
26 4
2 28 7
 


52 52 52 52 13
black ace A♣,♠
(e)
12 13 3 22
 

52 52 52 52
6
2
8
4



36 36 36 18
(b) 'Greater than 9' would be 10, 11, or 12. Add these probabilities:
3
2
1
6 1
 


36 36 36 36 6
(2) (a) 'or' with disjoint events. Just add:
(c) The 'given' means conditional probability, and the sample space S has been reduced
6 1

to the 18 odd rolls. There are 6 ways to roll a 7, so the probability would be:
18 3
(3) This is an 'and' question involving independent events, so we're going to multiply,
taking into account the changing number of balls as they're drawn.
5 4 20 5
5 4 3 60
5


(a) * 
(b) * * 
9 8 72 18
9 8 7 504 42
one less blue ball, one less ball
one less ball, one less red ball
(4) This, too, is an 'and' event, and we have to track the decreased number of cards with
each draw.
4 4 3
* *
52 51 50
one less King, two less cards
(5) Again, these are independent events with an 'and' condition. We multiply:
(a) (.22)3  .01 or 1 percent.
(b) This is the complement of 'none', so we take 1 - .01 = 99 percent.
(6) Just multiply the smaller paths which make up the bigger one: 3 * 4 = 12.
(7) Note the use of the word 'arrange' – this means we want a permutation:
8!
8  7  6  5  4  3!

 6560
8 P5 
(8  5)!
3!
(8) (a) There are 26 choices for each letter, 10 for each digit (if repeats are allowed), so:
263 *10 4 ways all told.
(b) 26  25  24 10  9  8  7 - the decreasing numbers reflect the removed choice (no
repeats).
(9)
52
C5
(10) (a) We have to use combinations (no rank on sub-committee) , together with the
multiplication principle: 8 C3 *6 C2
ways of choosing Dems
ways of choosing Repubs
(b) Here, the party (as stipulated) doesn't matter. We're simply choosing 3 from 14, and
order counts, so we have a permutation: 14 P3
(11) (a)
C5
52 C5
13
C3 *49 C2
52 C5
3 K's
Two other cards
(b)
4
C3 *4 C2
52 C5
three Aces, two kings
(c)
4
(12) This is a combination problem – we need to count the number of ways first of
selecting four people from the (cumulative) 14: 14 C4 . This will be the denominator of
our answer. Now we need to count:
C * C
(a) the number of ways of choosing 2 men , 2 women: 8 C2 *6 C2 , and divide: 8 2 6 2
14 C4
C
(b) The number of ways of choosing 4 men, and then divide: 8 4
14 C4