Given:
![\begin{gathered} Total\text{ number of crayons = 24} \\ Number\text{ of red crayons = 8} \\ Number\text{ of yellow crayons = 7} \\ Number\text{ of blue crayons = 9} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/qptp5g92pyreqczn5tqv4y417rlmmwh4wk.png)
Required:
Probability of removal of the crayon as blue, red, and yellow respectively
Step-by-step explanation:
The probability of removal of the first blue crayon is calculated as,
![\begin{gathered} Probability\text{ = }(^9C_1)/(^(24)C_1) \\ Probability\text{ =}(9)/(24) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/jar9oek5wsi6xrusmnfrxshx8vjl7gl1ew.png)
The probability of removal of the second red crayon is calculated as,
![\begin{gathered} Probability\text{ = }(^8C_1)/(^(24)C_1) \\ Probability\text{ = }(8)/(24) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/fxn8bi4z9ugz0tbxvbizhr5y2rckszq8tt.png)
The probability of removal of the third yellow crayon is calculated as,
![\begin{gathered} Probability\text{ = }(^7C_1)/(^(24)C_1) \\ Probability\text{ = }(7)/(24) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/xd1bn4mmoudnuiwy76wvyuddf9yionpxye.png)
Therefore the required probability is calculated as,
![\begin{gathered} Required\text{ probability = }(9)/(24)\text{ }*\text{ }(8)/(24)\text{ }*\text{ }(7)/(24) \\ Required\text{ probability = }(9*8*7)/(24*24*24) \\ Required\text{ probability = }(504)/(13824) \\ Required\text{ probability = 0.0365} \\ \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/l72hxsrxjyy38tq0femtenucuf4l5wgfy7.png)
Answer:
Thus the required probability is 0.0365