Question 1: The difference between the empirical probability, theoretical probability, and subjective probability
Empirical probability: This probability method is based on experiments. For instance, when a coin is tossed 1000 times, and we want to know how many times ahead will show up. When using the traditional way, the answer is considered to be 500. But when the empirical approach is used, an experiment will first be conducted whereby a coin will be tossed 1000 times, and then based on the observation from the experiment, the answer can be derived from there.
Theoretical probability: in this type of probability, there are certain numbers of events, for instance, “n” events and a basic probability formula is used to calculate the probability of the happening. For instance, when a coin is tossed once, the probability of getting a tail is ½. This is referred to as theoretical probability.
Subjective probability: this is an individual measure of a belief that an event will take place (it is based on a person’s intuition). It is rarely accurate and vague. For instance, an individual may have a feeling that on a specific day, there is a 60 percent probability that it will rain. This probability is purely based on the intuition of that individual because there is no formula which can be used to calculate.
Question 2: Odds in favor of winning the next game are 5:7; odds of tying the next game are 1:3; and the odds of losing the next game are 2:3. Can this prediction be correct?
This prediction is not correct.
The odds in favor of winning are 5:7, Therefore, p(winning) = 5/12
The odds in favor of tying are 1:3, Therefore, p(tying) =1/4
The odds in favor of losing are 2:3, Therefore, p(losing) the game = 2/5
The sum of the three probabilities is 64/60, which is not possible because the possible outcome of all the probabilities should be 1.
Question 3: If the odds against the Blue Jays winning this year’s World Series are 20:1, what is the probability that the Blue Jays will win this series?
Probability against winning= 20:1, therefore,
Odds for winning = 1:20, therefore, P(winning) = 1/21, or approximately 0.0476 or 4.8%
Question 4:What are the odds in favor of a total greater than 9 in a given roll of two standard dice?
Looking at all sums and how many times they occur,
2 = >1
2=>1
3=>2
4=>3
5=>4
6=>5
7=>6
8=>5
9=>4
10=>3
11=>2
12=>1
Therefore, a total greater than 9 would be 12, 10, or 11.
P(sum>9) = (1+3+2)/36 = 6/36 =1/6, therefore, the odds are 1:5
Question 5: a coin is tossed 4 times, what is the probability of tossing at least two heads?
The following are ways of tossing a coin four times:
No heads (1 count)
TTTT
1 head (4 counts)
HTTT
TTTH
THTT
TTHT
2 heads (6 counts)
HTTH
HHTT
TTHH
HTHT
THTH
THHT
3 Heads (4 counts)
THHH
HHHT
HTHH
HHTH
4 Heads (1 count)
HHHH
Total number of counts is 16
There are 6 instances of obtaining 2 heads. Therefore the probability is 6/16 = 3/8
Question 6: Two standard dice are rolled. What is the probability that a sum less than 6 is not rolled?
Possible outcomes for the two dice:
1 | 2 | 3 | 4 | 5 | 6 | |
1 | 1,1 | 1,2 | 1,3 | 1,4 | 1,5 | 1,6 |
2 | 2,1 | 2,2 | 2,3 | 2,4 | 2,5 | 2,6 |
3 | 3,1 | 3,2 | 3,3 | 3,4 | 3,5 | 3,6 |
4 | 4,1 | 4,2 | 4,3 | 4,4 | 4,5 | 4,6 |
5 | 5,1 | 5,2 | 5,3 | 5,4 | 5,5 | 5,6 |
6 | 6,1 | 6,2 | 6,3 | 6,4 | 6,5 | 6,6 |
Total number of outcomes = 36
The sum less than 6 rolled = 10/36,
Therefore the probability of a sum less than 6 not rolled = 1-10/36 = 26/36 = 13/18
Question 7: Suppose you randomly draw two marbles, without replacement, from a bag containing six green, four red, and three black marbles
- Determine the probability that both marbles are red
The bag contains 4 red marbles; therefore, one can possibly draw 4 reds from the bag which contains are a total of 13 marbles. Therefore, the probability of drawing a one red marble is 4/13
After getting a red from 4/13 chance, there was no replacement. Hence 3 reds are remaining in the bag.
Now the probability of getting the second red marble is 3/12, multiplying the two:
4/13 x 3/12 = 12/156 = 1/13
- Determine the probability that you pick at least green marble
Probability of picking at least one green = p(1 green) + p(2 greens)
= [(_{6}c_{1 }x_{ 7}c_{1}) + (_{6}c_{2 }+ _{7}c_{0})]/_{ 13}c_{2}
_{ }=(42 +15)/78
= 19/26
Question 8: Len just wrote a multiple-choice question with 15 questions, each having four choices. Len is sure that he got exactly 9 of the first 12 questions correct, but he guessed randomly on the last 3 questions. What is the probability that he will get at least 80% on the test?
To get a score of 80%, Len must get 12 questions right out of the total of 15 questions. If He got 9 questions out the 12 correct, in order to get a score of 80%, he must guess all of the remaining three questions right. Therefore,
P(3 right) = [1/4]^{3} = 1/64 or about 0.015625
Therefore, Len’s chance of getting 80% in the test is 1.6%
Question 9
For at least two teachers, the probability = 1 – probability of no teacher – probability of one teacher
= 1- []}/[]– []}/[]
= (300 – 84 – 630)/3003
= 2289/3003 or approximately 0.76
Question 10
There are _{10}P_{6 }ways of selecting the meal, if none of the friends pick the same entrée. Therefore the this event’s probability is,
_{10}P_{6}/10^{6 }= (10 x 9 x 8 x 7 x 6 x 5)/1000000
= 151200/1000000 = 0.1512
Hence the probability of at least two friends ordering the same entrée is 1 – 0.1512 = 0.8488, or84.9 percent.
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