Answer: ![\dfrac{2}{39}](https://tex.z-dn.net/?f=%5Cdfrac%7B2%7D%7B39%7D)
Step-by-step explanation:
Given : A mini bag of Skittles has 3 lemon, 4 grape, 4 orange and 2 lime Skittles.
Total Skittles = 3+4+4+2=13
Probability of pulling an orange Skittle out first : P(First orange) ![=\dfrac{\text{No. of orange skittles}}{\text{Total skittles}}](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B%5Ctext%7BNo.%20of%20orange%20skittles%7D%7D%7B%5Ctext%7BTotal%20skittles%7D%7D)
![=\dfrac{4}{13}](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B4%7D%7B13%7D)
Since the second skittles is drawn without replacement , so after drawing one skittle , the remaining skittles = 13-1=12
So , P(Second skittle is lime) =![\dfrac{\text{No. of lime Skittles}}{\text{remaining skittles }}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Ctext%7BNo.%20of%20lime%20Skittles%7D%7D%7B%5Ctext%7Bremaining%20skittles%20%7D%7D)
Since the events of pulling any Skittle are independent .
So , P( Orange then lime)= P(First orange) x P(Second skittle is lime)
![=\dfrac{4}{13}\times\dfrac{1}{6}=\dfrac{2}{39}](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B4%7D%7B13%7D%5Ctimes%5Cdfrac%7B1%7D%7B6%7D%3D%5Cdfrac%7B2%7D%7B39%7D)
∴ The probability of randomly pulling an orange Skittle out first and eating it and then pulling a lime Skittle out of the bag next = ![\dfrac{2}{39}](https://tex.z-dn.net/?f=%5Cdfrac%7B2%7D%7B39%7D)