Respuesta :
Answer:
[tex] P(A \cap B) = P(A) *P(B) =\frac{1}{5} \frac{1}{5}= \frac{1}{25}[/tex]
So then the best option would be:
a. 1/25
Step-by-step explanation:
For this case we assume that the sample space for the numbers is :
[tex] S_1= [A,B,C,D,E][/tex]
And the sample space for the numbers is:
[tex] S_2 =[1,2,3,4,5][/tex]
Both sampling spaces with a size of 5.
We define the following events:
A="We select a 2 from the numbers"
B= "We select a E from the letters"
We can find the individual probabilities for each event like this:
[tex] P(A)= \frac{1}{5}[/tex]
[tex] P(B) = \frac{1}{5}[/tex]
And assuming independence we can find the probability required like this:
[tex] P(A \cap B) = P(A) *P(B) =\frac{1}{5} \frac{1}{5}= \frac{1}{25}[/tex]
The last probability is the probability of obtain obtain a 2 AND an E
So then the best option would be:
a. 1/25
Answer: a. 1/25
P(2∩E) = 1/25
Step-by-step explanation:
Given;
The two set of numbers and letters
Set 1 = { 1,2,3,4,5}
Set 2 = { A,B,C,D,E}
The probability that he picks 2 and E equals the probability of picking 2 from set 1 multiplied by the probability of picking E from set 2.
Let P(2) represent the probability of picking 2 from set 1 ,
P(E) represent the probability of picking E from set 2 and
P(2∩E) represent the probability of picking 2 from set 1 and E from set 2.
P(2∩E) = P(2) × P(E) .....1
P(2) = N(2)/N(T)
N(2) = 1 (there is only one 2 in the set)
N(T) = 5 (total number of elements in set 1)
P(2) = 1/5
P(E) = N(E)/N(T)
P(E) = 1/5
Substituting P(E) and P(2) into equation 1
P(2∩E) = 1/5 × 1/5
P(2∩E) = 1/25
