Answer:
[tex]Probability = \frac{27}{80}[/tex]
Step-by-step explanation:
Given
[tex]Blue =9[/tex]
[tex]Yellow = 7[/tex]
Required
Determine the probability that one of the balls is blue while others is yellow
The order of selection may be any of these three:
Yellow, Yellow, Blue or Yellow, Blue, Yellow or Blue, Yellow, Yellow
Solving Yellow, Yellow, Blue
[tex]Probability = P(Y_1) * P(Y_2) * P(B)[/tex]
[tex]Probability = \frac{7}{16} * \frac{6}{15} * \frac{9}{14}[/tex]
[tex]Probability = \frac{1}{8} * \frac{1}{5} * \frac{9}{2}[/tex]
[tex]Probability = \frac{9}{80}[/tex]
Solving Yellow, Blue, Yellow
[tex]Probability = P(Y_1) * P(B) * P(Y_2)[/tex]
[tex]Probability = \frac{7}{16} * \frac{9}{15} * \frac{6}{14}[/tex]
[tex]Probability = \frac{9}{80}[/tex]
Solving Blue, Yellow, Yellow
[tex]Probability = P(B) *P(Y_1) * P(Y_2)[/tex]
[tex]Probability = \frac{9}{16} * \frac{7}{15} * \frac{6}{14}[/tex]
[tex]Probability = \frac{9}{80}[/tex]
Hence:
The required probability is:
[tex]Probability = \frac{9}{80} + \frac{9}{80} + \frac{9}{80}[/tex]
[tex]Probability = \frac{27}{80}[/tex]
