Answer:
[tex]P(White\ and\ Purple) = 2.52\%[/tex]
Step-by-step explanation:
Given
[tex]Purple = 18[/tex] [tex]Green = 34[/tex] [tex]White = 14[/tex]
[tex]Blue = 26[/tex] [tex]Black = 8[/tex]
Required
Probability of first selecting white and then selecting purple
The total boxes is:
[tex]Total = 18+34+14+26+8[/tex]
[tex]Total = 100[/tex]
The probability is represented as:
[tex]P(White\ and\ Purple)[/tex]
And the solution is:
[tex]P(White\ and\ Purple) = P(White) * P(Purple)[/tex]
[tex]P(White\ and\ Purple) = \frac{n(White)}{Total} * \frac{n(Purple)}{Total}[/tex]
[tex]P(White\ and\ Purple) = \frac{14}{100} * \frac{18}{100}[/tex]
[tex]P(White\ and\ Purple) = \frac{14*18}{100*100}[/tex]
[tex]P(White\ and\ Purple) = \frac{252}{10000}[/tex]
[tex]P(White\ and\ Purple) = 0.0252[/tex]
or
[tex]P(White\ and\ Purple) = 2.52\%[/tex]