Respuesta :
We will denote the first spinner as S5 and the second one as S6.
1) Probability spin the same number is both spinners twice
The probability of landing in a given number using S5 is equal to 1/5, while when using the S6 the probability is 1/6.
First, we get the same result twice using S5, this probability is given by:
[tex]P(S5_{\text{twice}})=\frac{1}{5}\cdot\frac{1}{5}=\frac{1}{25}[/tex](An specific number, of the 5 available, twice) Notice that the result we obtain with S5 does not affect what we obtain with S6.
On the other hand, the probability of getting any number twice in a row, using S6, is:
[tex]P(S6_{\text{twice}})=\frac{1}{6}\cdot\frac{1}{6}=\frac{1}{36}[/tex](An specific number, of the 6 available, twice) In case the problem refers to the probability of spinning S5 once, then S6, and obtaining the same number:
First, notice that there are a total of 5 results that satisfy this condition
(1,1),(2,2),(3,3),(4,4),(5,5)
And there is a total of 5*6=30 possible combinations. 30 different pairs (S5,S6).
So, the probability is the number of positive cases divided by the total amount of cases:
[tex]P(S5=S6)=\frac{5}{30}=\frac{1}{6}[/tex]This is the probability of getting the same number if you spin S5 and S6 once each.
2) Probability getting two numbers which SUM is equal to 5
Let's suppose that the problem refers to spinning once each one of the spinners and then adding the results.
First, we need to get the pairs that add up to 5
(S5,S6)
(1,4),(4,1)(2,3),(3,2). These are the only pairs that satisfy the condition.
And remember that, when spinning S5 and S6 once each, there are 30 possible combinations. So, the probability we are looking for in part 2 is:
[tex]P(SUM(5))=\frac{4}{30}=\frac{2}{15}\approx0.1333\ldots[/tex]3) Landing on an even number
In the case of S5, there are 2 even numbers:2,4 and 5 numbers on which the spinner can land:1,2,3,4,5.
So, the probability is:
[tex]P(S5_{\text{even}})=\frac{2}{5}=0.4[/tex]On the other hand, the probability of getting an even number with S6 is:
[tex]P(S6_{\text{even}})=\frac{3}{6}=0.5[/tex]We can even find the probability of spinning S5 once, then S6, and get an even number. Since the events are independent, that probability is:
[tex]P(S5_{\text{even}})\cdot P(S6_{\text{even}})=0.4\cdot0.5=0.2=\frac{1}{5}[/tex]d) Get one 2 and one 3.
Once again, there is a total of 2 pairs that satisfy this condition: (2,3) and (3,2), and there is a total of 30 combinations when we spin S5 and S6. So,
[tex]P(2and3)=\frac{2}{30}=\frac{1}{15}\approx0.0666[/tex]And that's the answer to the fourth question
