(3x5−2x4−5)−(2x4+x2−10) Subtract the two polynomials
2 answers:
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a) 1/6
b) 1/36
c)
1H, 2H, 3H, 4H, 5H, 6H
1T, 2T, 3T, 4T, 5T, 6T
Step-by-step explanation:
a)
The probability of a certain event A to occur is given by
where
a is the number of successfull outcomes, in which event A occurs
n is the total number of possible outcomes
In this problem, the event is
"getting a 6 when throwing a dice once"
We know that the possible outcomes of a dice are six: 1, 2, 3, 4, 5, 6, so we he have
The successfull outcome in this case is only if we get a 6, so only 1 outcome, therefore
So, the probability of this event is
b)
In this case instead, we are throwing the dice twice.
The two throws of the dice are independent events (one does not depend on the other): the probability that two independent events A and B occur at the same time is given by the product of the individual probabilities,
where
is the probability that event A occurs
is the probability that event B occurs
Here we have:
- Event A is "getting a 6 in the first throw of the dice". We already calculated this probability in part a), and it is
- Event B is "getting a 6 in the second throw of the dice". Since the dice has not changed, the probability is still the same, so
Therefore, the probability of getting a 6 on both throws is:
c)
In this problem, we have:
- A dice that is thrown once
- A coin that is also thrown once
The dice has 6 possible outcomes, as we stated in part a):
1, 2, 3, 4, 5, 6
While the coin has two possible outcomes:
H = head
T = tail
So, in order to find all the outcomes of the two events combined, we have to combine all the outcomes of the dice with all the outcomes of the coin.
Doing so, using the following notation:
1H (getting 1 with the dice, and head with the coin)
The possible outcomes are:
1H, 2H, 3H, 4H, 5H, 6H
1T, 2T, 3T, 4T, 5T, 6T
So, we have a total of 12 possible outcomes.
