M230 Test I Fall 1997Solutions

1. tex2html_wrap_inline84, tex2html_wrap_inline86 tex2html_wrap_inline88 tex2html_wrap_inline90
OR
tex2html_wrap_inline92, tex2html_wrap_inline94}, tex2html_wrap_inline88 tex2html_wrap_inline98

2. using combinations
(a)C(10,4)

(b)C(3,1)C(2,1)C(4,1)C(1,1)

(c) C(5,1)C(5,3)+C(5,2)C(5,2)+C(5,3)C(5,1)

(d)tex2html_wrap_inline106

using permutations
(a)P(10,4)

(b)P(3,1)P(2,1)P(4,1)P(1,1)P(4,4)

(c)C(5,1)C(5,3)P(4,4) + C(5,2)C(5,2)P(4,4) + C(5,3)C(5,1)P(4,4)

(d)tex2html_wrap_inline114

3.(a) tex2html_wrap_inline116
tex2html_wrap_inline118 so there are tex2html_wrap_inline120 elements in tex2html_wrap_inline122.

(b)tex2html_wrap_inline124 tex2html_wrap_inline126 so there are
tex2html_wrap_inline128 elements in tex2html_wrap_inline124.

(c)tex2html_wrap_inline132 tex2html_wrap_inline134} so there are tex2html_wrap_inline136 elements
in tex2html_wrap_inline132.

(d)tex2html_wrap_inline140 tex2html_wrap_inline142 so the number of elements in tex2html_wrap_inline140 is 4 + 3 = 7.

4. Let x, y, z, and w be the probability of the event indicated on the Venn diagram below.

First we will find the
values of x, y, z,
and w, and then relate
them with the probabilities
that we need to find.

x + y = .45 y + z = .70 y = .25 w=1-x-y-z
so x = .20, z = .45, and w=.10

(a)tex2html_wrap_inline178 (b)tex2html_wrap_inline180

(c)tex2html_wrap_inline182 (d)tex2html_wrap_inline184

5. For this problem, the following symbols will be used to represent the suits of a deck of cards:
Hearts = tex2html_wrap_inline186 Diamonds = tex2html_wrap_inline188 Clubs = tex2html_wrap_inline190 Spades = tex2html_wrap_inline192

There are tex2html_wrap_inline194 ways to pick the 3 suits to appear in the hand. We will look at the choices for the case where we have tex2html_wrap_inline198 and then multiply that number by 4.

the possibilities are: The number of choices are:
tex2html_wrap_inline202 C(13,2)C(13,2)C(13,1)
tex2html_wrap_inline206 C(13,2)C(13,1)C(13,2)
tex2html_wrap_inline210 C(13,1)C(13,2)C(13,2)
tex2html_wrap_inline214 C(13,1)C(13,3)C(13,1)
tex2html_wrap_inline218 C(13,1)C(13,1)C(13,3)
tex2html_wrap_inline222 C(13,3)C(13,1)C(13,1)

Add all of these cases together to get 3 [ C(13,2)C(13,2)C(13,1) + C(13,1)C(13,1)C(13,3) ].

remember that each of the 3 suits must appear once in the 5-card hand!

So the answer is 4(3 [ C(13,2)C(13,2)C(13,1) + C(13,1)C(13,1)C(13,3) ] ).

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Author: Rebecca Sullivan
Fri Jan 29 15:22:29 EST 1999