A fair tetrahedral dice (with face number 1,2,3,4)and a fair coin are tossed together.
A) construct a table for the sample space of the random experiment.
B) using your sample space. Find the probability that:
a) A tail and an odd number show up
b) A head and a square show up
c) A head show up
A) Sample Space:
Toss 1 | Toss 2 | Result
Head | 1 | Head, 1
Head | 2 | Head, 2
Head | 3 | Head, 3
Head | 4 | Head, 4
Tail | 1 | Tail, 1
Tail | 2 | Tail, 2
Tail | 3 | Tail, 3
Tail | 4 | Tail, 4
B) a) P(Tail, Odd) = P(Tail, 1) + P(Tail, 3) = 1/4 + 1/4 = 1/2
b) P(Head, 4) = 1/4
c) P(Head) = P(Head, 1) + P(Head, 2) + P(Head, 3) + P(Head, 4) = 1/2
A) Here is a table representing the sample space for the random experiment:
| | Head | Tail |
|---------------|------|------|
| 1 (tetrahedron)| H1 | T1 |
| 2 (tetrahedron)| H2 | T2 |
| 3 (tetrahedron)| H3 | T3 |
| 4 (tetrahedron)| H4 | T4 |
B) Let's find the probabilities:
a) The probability of getting a tail and an odd number can be calculated by finding the number of favorable outcomes (tail and odd number) and dividing it by the total number of outcomes.
Number of favorable outcomes: 2 (T1 and T3)
Total number of outcomes: 8 (4 tetrahedron faces x 2 coin sides)
Probability = favorable outcomes / total outcomes = 2/8 = 1/4
b) The probability of getting a head and a square number can be calculated in a similar way:
Number of favorable outcomes: 2 (H4 and H2)
Total number of outcomes: 8 (4 tetrahedron faces x 2 coin sides)
Probability = favorable outcomes / total outcomes = 2/8 = 1/4
c) The probability of getting a head can be calculated by counting the number of outcomes with heads and dividing it by the total number of outcomes:
Number of outcomes with heads: 4 (H1, H2, H3, H4)
Total number of outcomes: 8 (4 tetrahedron faces x 2 coin sides)
Probability = outcomes with heads / total outcomes = 4/8 = 1/2
So, the probabilities are:
a) 1/4
b) 1/4
c) 1/2
A) The sample space for the random experiment can be constructed using a table, where the outcomes are listed for each possible combination of the dice and coin:
| Dice | Coin |
|:------:|:----:|
| 1 | Head |
| 1 | Tail |
| 2 | Head |
| 2 | Tail |
| 3 | Head |
| 3 | Tail |
| 4 | Head |
| 4 | Tail |
B) Probability calculations using the sample space:
a) To find the probability of getting a tail and an odd number showing up, we look for outcomes in which the Coin is a Tail and the Dice number is odd.
There are two outcomes: (1, Tail) and (3, Tail) out of a total of eight outcomes.
Therefore, the probability is 2/8, which simplifies to 1/4.
b) To find the probability of getting a head and a square showing up, we look for outcomes in which the Coin is a Head and the Dice number is a square (i.e., 1 or 4).
There are two outcomes: (1, Head) and (4, Head) out of a total of eight outcomes.
Therefore, the probability is 2/8, which simplifies to 1/4.
c) To find the probability of getting a head showing up, we count the number of outcomes in which the Coin is a Head.
There are four outcomes where the Coin is a Head: (1, Head), (2, Head), (3, Head), and (4, Head).
Therefore, the probability is 4/8, which simplifies to 1/2.
A) To construct a table for the sample space of the random experiment, we need to list all possible outcomes when tossing a fair tetrahedral dice and a fair coin together.
The tetrahedral dice has 4 sides, numbered 1, 2, 3, and 4.
The coin has 2 sides, which we can label as "H" for heads and "T" for tails.
Let's list all possible outcomes by combining the results of both the dice and the coin:
1. (1, H)
2. (1, T)
3. (2, H)
4. (2, T)
5. (3, H)
6. (3, T)
7. (4, H)
8. (4, T)
So, the sample space is:
{(1, H), (1, T), (2, H), (2, T), (3, H), (3, T), (4, H), (4, T)}.
B) Now, let's use the sample space to find the probability of the following events:
a) A tail and an odd number show up:
The outcomes that fulfill this event are (1, T), (3, T).
So, the probability is 2/8 = 1/4.
b) A head and a square show up:
The outcomes that fulfill this event are (1, H), (4, H).
So, the probability is 2/8 = 1/4.
c) A head shows up:
The outcomes that fulfill this event are (1, H), (2, H), (3, H), (4, H).
So, the probability is 4/8 = 1/2.