Respuesta :
#1) 5/20
#2) 7/300
#3) 3/50
#4) 5/20
Explanation
For #1:
There are 15 even numbers out of 30. Since it is replaced before drawing the second ball, there will be 15 odd numbers out of 30. This gives us
15/30(15/30) = 225/900 = 5/20.
For #2:
There is 1 7 out of 30; then there are 14 numbers greater than 16 out of 30:
1/30(14/30) = 14/900 = 7/300
For #3:
There are 6 multiples of 5 out of 30; then there are 9 prime numbers out of 30:
6/30(9/30) = 54/900 = 3/50
For #4:
There are 15 even numbers out of 30; then there are still 15 even numbers out of 30:
15/30(15/30) = 225/900 = 5/20
#2) 7/300
#3) 3/50
#4) 5/20
Explanation
For #1:
There are 15 even numbers out of 30. Since it is replaced before drawing the second ball, there will be 15 odd numbers out of 30. This gives us
15/30(15/30) = 225/900 = 5/20.
For #2:
There is 1 7 out of 30; then there are 14 numbers greater than 16 out of 30:
1/30(14/30) = 14/900 = 7/300
For #3:
There are 6 multiples of 5 out of 30; then there are 9 prime numbers out of 30:
6/30(9/30) = 54/900 = 3/50
For #4:
There are 15 even numbers out of 30; then there are still 15 even numbers out of 30:
15/30(15/30) = 225/900 = 5/20
Answer:
Given : A bag contains 30 lottery balls numbered 1-30 a ball is selected replaced then another is drawn.
To find : Each probability
1) p ( and even,then odd )
2) p ( 7, then a number greater than 16)
3) p ( a multiple of 5, then a prime number )
4) p ( two even number )
Solution :
[tex]\text{Probability}=\frac{\text{Favorable outcome}}{\text{Total number of outcome}}[/tex]
1) There are 15 even numbers and 15 odd numbers.
Probability of getting even first then odd is
[tex]\text{P(even,then odd)}=\frac{15}{30}\times\frac{15}{30}[/tex]
[tex]\text{P(even,then odd)}=\frac{225}{900}=\frac{1}{4}[/tex]
2) Number greater than 16 out of 30 are 14.
Probability of getting 7 first then a number greater than 16 is
[tex]\text{P(7, then a number greater than 16)}=\frac{1}{30}\times\frac{14}{30}[/tex]
[tex]\text{P(7, then a number greater than 16)}=\frac{14}{900}=\frac{7}{450}[/tex]
3) Multiple of 5 - 5,10,15,20,25,30=6
Prime numbers - 2,3,7,9,11,13,17,19,23,29=10
Probability of getting a multiple of 5, then a prime number is
[tex]\text{P(a multiple of 5, then a prime number )}=\frac{6}{30}\times\frac{10}{30}[/tex]
[tex]\text{P(a multiple of 5, then a prime number )}=\frac{60}{900}=\frac{1}{15}[/tex]
4) There are 15 even numbers.
Probability of getting two even number is
[tex]\text{P( two even number)}=\frac{15}{30}\times\frac{15}{30}[/tex]
[tex]\text{P( two even number)}=\frac{225}{900}=\frac{1}{4}[/tex]
