MATH 111 Homework 8: Probability Solutions and Comments, Assignments of Trigonometry

Solutions and comments for various probability problems in math 111 homework 8. Topics include calculating probabilities of heads in coin flips and die rolls, finding probabilities of specific outcomes in multiple trials, and understanding events and outcomes in probability experiments.

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Pre 2010

Uploaded on 08/26/2009

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MATH 111: HW 8 SOLUTIONS AND COMMENTS
7.1, #3. Since 53 out of 100 tosses came up heads, 100 53 = 47 of them came up tails.
So, 47 out of 100, or 47%, of the tosses were tails.
7.1, #7. In case you’re curious, 1572 of the class’ 3100 spins came up heads, or approxi-
mately 50.7%.
7.1, #9. The king was found in 491 of the class’ 1250 trials, which is approximately 39%
of the trials.
7.2, #1. The probability that he wears the black shoes is 1 3
5=2
5.
7.2, #2. There are four equally likely possible outcomes:
Dime Penny
H H
H T
T H
T T
Of these outcomes, the first three (HH, HT, TH) comprise the event that at least one head
is flipped. Therefore, the probability of this event is 3/4.
7.2, #3. There are a total of 14 chocoloates, each of which was equally likely to have
been picked. Therefore, the probability of picking a green chocolate was 2/14 = 1/7. The
probability of picking a blue chocolate was 5/14.
I like chocolate, so the probability that I would eat the chocolate is 1. This probability may
or may not be different for you!
7.2, #13. There are 6 equally likely outcomes of the die roll, and 2 equally likely outcomes
of the coin flip, so there are 6×2 = 12 equally likely outcomes for this experiment (in which a
die is rolled and independently a coin is flipped). The event that Roosevelt can be seen (i.e.,
the coin lands heads) and 4 is rolled is one of these 12 outcomes; therefore, the probability
of this event is 1/12.
If we’re told that the die roll was 2, then there are only two remaining possible equally likely
outcomes: 2 with heads; and 2 with tails. The dime lands tails in one of these two outcomes,
so the probability that the dime lands tails under these circumstances is 1/2.
7.2, #18. There are 38 equally likely outcomes (0, 00, and 1 through 36). The probability
of the single outcome 13 is therefore 1/38.
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MATH 111: HW 8 SOLUTIONS AND COMMENTS

7.1, #3. Since 53 out of 100 tosses came up heads, 100 − 53 = 47 of them came up tails. So, 47 out of 100, or 47%, of the tosses were tails.

7.1, #7. In case you’re curious, 1572 of the class’ 3100 spins came up heads, or approxi- mately 50.7%.

7.1, #9. The king was found in 491 of the class’ 1250 trials, which is approximately 39% of the trials.

7.2, #1. The probability that he wears the black shoes is 1 − 35 = 25.

7.2, #2. There are four equally likely possible outcomes:

Dime Penny H H H T T H T T

Of these outcomes, the first three (HH, HT, TH) comprise the event that at least one head is flipped. Therefore, the probability of this event is 3/4.

7.2, #3. There are a total of 14 chocoloates, each of which was equally likely to have been picked. Therefore, the probability of picking a green chocolate was 2/14 = 1/7. The probability of picking a blue chocolate was 5/14.

I like chocolate, so the probability that I would eat the chocolate is 1. This probability may or may not be different for you!

7.2, #13. There are 6 equally likely outcomes of the die roll, and 2 equally likely outcomes of the coin flip, so there are 6×2 = 12 equally likely outcomes for this experiment (in which a die is rolled and independently a coin is flipped). The event that Roosevelt can be seen (i.e., the coin lands heads) and 4 is rolled is one of these 12 outcomes; therefore, the probability of this event is 1/12.

If we’re told that the die roll was 2, then there are only two remaining possible equally likely outcomes: 2 with heads; and 2 with tails. The dime lands tails in one of these two outcomes, so the probability that the dime lands tails under these circumstances is 1/2.

7.2, #18. There are 38 equally likely outcomes (0, 00, and 1 through 36). The probability of the single outcome 13 is therefore 1/38.

The event that the spin lands on red consists of 18 of the outcomes (half of the outcomes 1 through 36). The probability of this event is therefore 18/36 = 9/18.

7.2, #21. Of the 36 equally likely rolls of a pair of dice, 6 are doubles; therefore, the probability of this event is 6/36 = 1/16.

7.2, #27. There are eight equally likely possible outcomes of three coin flips: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. However, we’ve been told that the coin flips were not all tails, so TTT is not a possibility. This leaves seven equally likely possible outcomes: HHH, HHT, HTH, HTT, THH, THT, TTH. The probability that all the flips were heads is just the probability of the outcome HHH; this is one out of seven equally likely outcomes, so the probability is 1/7.