A fair tetrahedral die (with face 1,2,3 and 4)and a fair coin are tosses together
Construct a table of sample space of a random experiment
Use your sample space to.find the probability that
(1):a tail and an odd number show up
(2):a head and a square show up
(3): a head show up
plz help one Mr obleck gave me a sample space but am still confused....
I want solution to learn
Can't find the previous response by Obleck, but he probably had listed these:
H1
H2
H3
H4
T1
T2
T3
T4
Now, simply count how many are tail and odd: T1, T3
out of how many ? ---- 8
so prob(tail and odd) = 2/8 = 1/4
how many are: head and a square show up ---> H1, H4
etc.
surely you can do this!
oooh you beauty thannnnnnnnnnk you so much much much .....
Sure! Let's break this problem down step by step.
First, let's construct the sample space for this random experiment. The sample space is simply the set of all possible outcomes. In this case, we have two distinct events: the toss of the die and the toss of the coin.
For the die toss, the possible outcomes are the numbers on the die, which are 1, 2, 3, and 4.
For the coin toss, the possible outcomes are heads and tails.
To construct the sample space, we consider all possible combinations of the outcomes of the die toss and the coin toss. Here's a table to represent the sample space:
| T | H
------------------
1 | 1T | 1H
2 | 2T | 2H
3 | 3T | 3H
4 | 4T | 4H
The table shows all possible outcomes of the random experiment. For example, 2T means that the die shows a 2 and the coin shows a tail.
Now, let's find the probabilities.
(1) To find the probability that a tail and an odd number show up, we need to find all the outcomes where a tail and an odd number occur together. From the table, we can see that this happens in two cases: 1T and 3T. So, there are 2 favorable outcomes.
Since the sample space consists of 8 equally likely outcomes (4 numbers on the die multiplied by 2 coin toss outcomes), the probability can be calculated as favorable outcomes divided by total outcomes: 2/8 = 1/4.
(2) To find the probability that a head and a square show up, we need to find all the outcomes where a head and a square number occur together. From the table, we can see that this happens only in one case: 4H. So, there is 1 favorable outcome.
Again, with 8 equally likely outcomes, the probability is 1/8.
(3) Finally, to find the probability that a head shows up, we need to find all the outcomes where a head occurs. From the table, we can see that this happens in four cases: 1H, 2H, 3H, and 4H. So, there are 4 favorable outcomes.
With 8 equally likely outcomes, the probability is 4/8 = 1/2.
I hope this helps! Let me know if you have any more questions.