Answer:
Probability of drawing two yellow marbles if the first one is placed back before the second draw is 1 : 5.
Step-by-step 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.
[tex]P(A) = \frac{\text {event occurred}}{\text{ total outcomes}}[/tex]
[tex]P(A) = \frac{2}{10}[/tex]
Similarly
[tex]P(B) = \frac{\text {event occurred}}{\text{ total outcomes}}[/tex]
[tex]P(B) = \frac{2}{10}[/tex]
Since the first one is placed back before the second draw.
Now P(A) and P(B) are independent event
[tex]P(A \text{ and }B) = P(A) \cdot P(B)[/tex]
=> [tex]\frac{2}{10} \times \frac{2}{10}[/tex]
=>[tex]\frac{4}{20}[/tex]
=> [tex]\frac{2}{10}[/tex]
=> [tex]\frac{1}{5}[/tex]
