Answer:
[tex]\dfrac{16}{225}[/tex].
Step-by-step explanation:
It is given that,
Number of red markers = 3
Number of blue markers = 5
Number of yellow markers = 3
Number of black markers = 4
Total markers = 3 + 5 + 3 + 4 = 15
Probability of getting a black marker is
[tex]P(Black)=\dfrac{\text{Number of black markers}}{\text{Total markers}}[/tex]
[tex]P(Black)=\dfrac{4}{15}[/tex]
Probability of choosing two black markers if the first marker is put back before the second is drawn, is
Required probability [tex]=P(Black)\times P(Black)[/tex]
[tex]=\dfrac{4}{15}\times \dfrac{4}{15}[/tex]
[tex]=\dfrac{16}{225}[/tex]
Therefore, the probability of choosing two black markers if the first marker is put back before the second is drawn, is [tex]\dfrac{16}{225}[/tex].
