A bag contains 3 purple and 5 orange jelly-beans. If one jelly-bean is selected and is NOT replaced and a second jelly-bean is selected, find P(orange and then a purple) *

A bag contains 3 purple and 5 orange jelly-beans. If one jelly-bean is selected and is NOT replaced and a second jelly-bean is selected, find P(two purples in a row) *

P(orange and then a purple) is 15/56.

P(two purples in a row) is 6/56.

Step-by-step explanation:

Without replacement, find P(orange and then a purple) :

P (orange) = No.of orange jelly-beans / total jelly-beans.

⇒ 5 / 8

P (purple) = No.of purple jelly-beans / remaining jelly-beans from the total

⇒ 3 / 7

P(orange and then a purple) = P (orange) × P (purple)

⇒ 5/8 × 3/7

⇒ 15/56

Without replacement, find P(two purples in a row) :

P (1st purple) = No.of purple jelly-beans / total jelly-beans.

⇒ 3 / 8

P (2nd purple) = remaining no.of purple jelly-beans / remaining jelly-beans from the total.

⇒ 2 / 7

P(two purples in a row) = P (1st purple) × P (2nd purple)

⇒ 3/8 × 2/7

⇒ 6/56

i used to know how to do this in 7th but i forget easily

step-by-step explanation:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

answer:

f(5/2) = 10/3

i hope this you!