Answer:
0.9999
Step-by-step explanation:
Each bag has a total of 20 + 15 + 5 + 10 = 50 candies.
Each of the 15 kids get their own back and randomly pick 12 candies (without replacement)
First we are going to calculate the probability that they don't get any green candy.
P (not getting a green candy) = C₄₀,₁₂ /C₅₀,₁₂ .
Therefore, the probability of getting at least one candy is 1 - P(not getting a green candy)
P (getting at least one candy) = 1 - (C₄₀,₁₂/C₅₀,₁₂) = 1 - 0.0460 = 0.954
So the probability that one kid gets at least one green candy is 0.954
But now we need to find the probability that at least two of the 15 kids will each have at least one green candy.
To do this, we can use a binomial distribution where success will be getting at least one green candy with p = 0.954
Fail will be not getting a green candy with p = 1 -0.954 = 0.046.
k = 15
We need at least two kids to have success, so we can do P(at least 2 kids have success) = 1 - (P( X=0) + P (X=1))
P (X=0) == <u>8.737103395697172336050176 × 10⁻²¹</u>
P(X=1) = = <u>2.71799890418318556801908736 × 10⁻¹⁸</u>
Therefore:
P(at least 2 kids have success) = 1 - (P( X=0) + P (X=1))
= 1 - (8.737103395697172336050176 × 10⁻²¹ + 2.71799890418318556801908736 × 10⁻¹⁸) = 0.9999