5. Consider the probabilities in problem 4. The probability that a randomly selected single
book purchase will not be for a work of fiction is
(a) 0.80
(b) 0.25
(c) 0.20
(d) 0.30
(e) 0.75
6. Two dice are rolled. Suppose A is the event the sum of the numbers on the top faces is
even, and B is the event that one of the dice shows the number “1.” The probability of
event B is
(a) 1/2
(b) 18/36
(c) 12/36
(d) 11/36
(e) 1/3
3
7. A child is playing with three dice: red, green, and blue. The probability that she will roll a
5 on the red, a 4 on the green, and a 3 on the blue is:
(a) 1/2
(b) 1/36
(c) 1/216
(d) 4/36
(e) 1/3
(6) 11/36
(7) 1/216
6. No, either-or probabilities are found by adding the individual probabilities.
P = P(1) + P(1) = ?
7. If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events.
Well, there are 11 possible combinations that one of two dices rolls 1.
i.e. (1,1) (1,2),....(2,1), (3,1)....
probability for single event P(1)= 1/36.
then Adding all 11 events is
P = P(1) + P(1) +P(1) +...... 11 times
= 1/36 + 1/36 + 1/36 +.....
=11/36.
additionally probability for event A is 18/36
Consider each die separately.
The P for die 1 to have a 1 = 1/6
The P for die 2 to have a 1 = 1/6
You want either one or the other of the dice to show a 1. Either-or probabilities are found by adding the individual probabilities.
rolling two dice together, what is the probability of getting (6,6)?
To find the probability that a randomly selected single book purchase will not be for a work of fiction, we need to consider the probabilities from problem 4.
In problem 4, we were given the following probabilities:
- The probability of buying a work of fiction was 0.40 (Let's call this probability A)
- The probability of buying a non-fiction book was 0.60 (Let's call this probability B)
Now, to find the probability that a book purchase will not be for a work of fiction, we can subtract the probability of buying a work of fiction from 1 (since the sum of probabilities for all possible outcomes is always 1).
So, the probability of not buying a work of fiction = 1 - Probability of buying a work of fiction
= 1 - A
= 1 - 0.40
= 0.60
Therefore, the answer is (c) 0.20.
To find the probability of event B in problem 6, where event B is that one of the dice shows the number "1", we need to consider the possible outcomes.
When two dice are rolled, there are 36 possible outcomes (each die has 6 possible outcomes).
Out of these 36 outcomes, the number of outcomes where at least one die shows the number "1" can be found by subtracting the number of outcomes where no die shows the number "1" from the total number of outcomes.
The number of outcomes where no die shows the number "1" is 5 * 5 = 25 (since for each die, there are 5 possible outcomes other than 1).
So, the number of outcomes where at least one die shows the number "1" is 36 - 25 = 11.
Therefore, the probability of event B is the number of outcomes where event B occurs divided by the total number of outcomes.
Probability of event B = Number of outcomes where event B occurs / Total number of outcomes
= 11 / 36
Hence, the answer is (d) 11/36.
In problem 7, where a child is rolling three dice, we need to find the probability that she will roll a 5 on the red die, a 4 on the green die, and a 3 on the blue die.
The probability of rolling a 5 on one die is 1/6, since there are 6 possible outcomes and only 1 of them is a 5.
Similarly, the probability of rolling a 4 on one die is also 1/6.
And the probability of rolling a 3 on one die is 1/6.
To find the probability of all three events happening, we multiply the probabilities of each event together.
Probability of rolling a 5 on the red die, a 4 on the green die, and a 3 on the blue die = Probability of rolling a 5 on the red die * Probability of rolling a 4 on the green die * Probability of rolling a 3 on the blue die
= (1/6) * (1/6) * (1/6)
= 1/216
Therefore, the answer is (c) 1/216.