Question

A mini bag of Skittles has 3 lemon, 4 grape, 4 orange and 2 lime Skittles. What is the probability of randomly pulling an orange Skittle out first and eating it and then pulling a lime Skittle out of the bag next?

Answer 1

I think that the answer would be 2 out of 36 chance or 1 in 18 chance.

Answer 2

Answer: $\dfrac{2}{39}$

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) $=\dfrac{\text{No. of orange skittles}}{\text{Total skittles}}$

$=\dfrac{4}{13}$

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) =$\dfrac{\text{No. of lime Skittles}}{\text{remaining skittles }}$

$=\dfrac{2}{12}=\dfrac{1}{6}$

Since the events of pulling any Skittle are independent .

So , P( Orange then lime)= P(First orange) x P(Second skittle is lime)

$=\dfrac{4}{13}\times\dfrac{1}{6}=\dfrac{2}{39}$

∴ The probability of randomly pulling an orange Skittle out first and eating it and then pulling a lime Skittle out of the bag next = $\dfrac{2}{39}$