| Finite Mathematics |
Test #2
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April 25, 2001
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R. Hammack
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| Name: ________________________ |
Score: _________
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From the following Venn diagram, you can see that 90,000 customers have neither service.
_________________________________________ U _____________ | ______________________ | | | | 90,000 | | | _________|_____________________ | | | 50,000 | | | | | | | 20,000 | | | | |___ F _____|________| 40,000 | | | |_____________________ W _____| | |_______________________________________________________|
(2) A club consists of 60 men and 40 women. To fairly choose a president and a secretary, names of all members are put into a hat and two names are drawn. The first name drawn is the president, and the second name drawn is the secretary. What is the probability that the president and the secretary have the same gender?
The sample space is S = {mm, mf, fm, ff} and we are looking for the probability of the event E = {mm, ff}. To find the probabilities of mm and ff, we can make a probability tree.
m P(mm) = (60/100)(59/99)
/
59/99/
/
/
m
/ \
60/100 / \
/ \
/ \
/ f
\ m
\ /
\ /
40/100 \ /
\ /
f
\
\
39/99 \
\
f P(ff) = (40/100)(39/99)
The probability we seek is P({mm, ff}) = P(mm) + P(ff) = (60/100)(59/99) +
(40/100)(39/99) ~ 51.51%
(3) You toss a fair coin 8 times. What is the probability that you do not get 4 heads?
The sample space S = {HHHHHHH, HHHHHHHT, HHHHHHTH, ... } consists of all sequences
of 8 H's and T''s,
so n(S) = (2)(2)(2)(2)(2)(2)(2)(2) = 256.
We seek P(E), where the event E consists of those outcomes in S not having exactly 4 heads.
The complement E' of E consists of all outcomes in S with exactly 4 heads, and that's an easier event to work with.
By the complement formula, P(E) = 1 - P(E') = 1 - n(E')/n(S) = 1 - n(E')/256.
Thus, to find P(E) we just need to compute the number n(E').
To make an event in E' you choose 4 out of 8 slots for an H, then fill in the
rest of the slots with T's.
For example, _ _ H _ HH_H becomes TTHTHHTH
There are C8, 4 = 8!(4!(8 - 4)!) = 8!/(4!)(4!) = (8 7 6 5)/(4 3 2 1) = 70 ways to do this, so n(E') = 70
Thus P(E) = 1 - P(E') = 1 - n(E')/n(S) = 1 - n(E')/256 = 1 - 70/256 = 186/256 = 93/128 ~ 72.65%
(4) There is a 40% chance of rain on Saturday and a 25% chance of rain on
Sunday. What is the probability that it will rain on the weekend? (You may assume
that the events "Rain on Saturday" and "Rain on Sunday"
are independent.)
A = "Rain on Saturday"
B = "Rain on Sunday"
We want P(A and B).
P(A and B) = P(A) + P(B) - P(A∩B) = P(A) + P(B) - P(A)P(B) = .40 + .25 -
(.40)(.25) = .55
So the chance of rain over the weekend is 55%.
(5) A four-card hand is dealt off a shuffled 52-card deck. What is the probability that the cards are all of the same color (i.e. all red or all black)?
Let E be the event "cards have the same color," so we seek P(E)
There are C26, 4 ways to get two red cards and C26, 4
ways to get two black cards.
Thus there are C26, 4 + C26, 4 = 2 C26, 4 ways
to get a hand with two cards of the same color.
Therefore P(E) = n(E)/n(S) = 2C26, 4 / C52,4 ~ 11.04%
(6) Two cards are dealt off a shuffled 52-card deck.What is the probability
that the cards are both red or both aces?
A = "Both red"
B = "Both Aces"
We seek P(A or B) = P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
Now, A∩B is the hand consisting of the 2 red aces, so n(A∩B) = 1
Thus P(A ∪ B)
= P(A) + P(B) - P(A ∩ B)
= C26,2 / C52, 2 + C4,2 / C52, 2
- 1 / C52, 2
= 325/1326 + 6/1326 - 1/1326
= 330/1326 ~ 24.88%
(7) A card is drawn off a 52-card deck. Let A be the event "the card is a heart." Let B be the event "the card is a queen."
(a) Are events A and B independent or dependent? Explain.
P(A) = 13/52 = 1/4
Now if B has happened, the card drawn is a queen, and theres still a one in 4 chance that it is the queen of hearts, so P(A|B) = 1/4
Therefore, since P(A) = P(A|B), the events A and B are independent.
(b) Are events A and B mutually exclusive? Explain.
No, because their intersection is the queen of hearts, not the empty set.
| (8) A box contains 6 lettered blocks, as illustrated. You randomly draw out two blocks, one after the other. |
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(a) What is the probability that both draws are vowels?
E = "First draw is a vowel"
F = "Second draw is a vowel"
P(E and F) = P(E)P(F|E) = (2/6)(1/5) = 1/15 ~ 6.66%
(b) What is the probability that the first draw is a U?
1/6 ~ 16.66%
(c) What is the probability of the event E = "both draws are vowels or the first draw is a U?"
A = "both vowels"
B = "First is a U"
P(A or B) = P(A) + P(B) - P(A and B) = 1/15 + 1/6 - 1/30 = 2/30 + 5/30 -1/30 = 6/30 = 1/5 ~ 20%
Note, P(A and B) = 1/30 is computed in part d.
(d) What is the probability that both draws are vowels and the first draw is a U?
A = "both vowels"
B = "First is a U"
P(A and B) = P(A)P(B|A) = (1/15)(1/2) = 1/30 ~ 3.33%