Answer:
Probability of drawing two yellow marbles if the first one is placed back before the second draw is 1 : 5.
Explanation:
Given:
Number of yellow marbles = 2
Number of pink marbles = 3
Number of blue marbles = 5
Total number of marbles = 10
To Find:
Probability of drawing two yellow marbles if the first one is placed back before the second draw = ?
Solution:
Let
P(A) be the probability of drawing the first marble
P(B) be the probability of drawing the second marble.


Similarly


Since the first one is placed back before the second draw.
Now P(A) and P(B) are independent event

=>

=>

=>

=>
