A cupcake stand has 40 chocolate, 30 coconut and 20 banana cupcakes. Alice chooses 20 cupcakes at random to create a box as a present for her friend.
What is the probability that she chose:
(a) Eight banana and 6 coconut cupcakes?
(b) At least 2 chocolate cupcakes?
(c) All cupcakes of the same kind?

This is about permutations and combinations.

a) Probability = 0.00563

b) Probability = 0.99997522

c) Probability = 0.0000000027045

Chocolate = 40

Coconut = 30

Banana = 20

Total number of chocolates = 40 + 30 + 20

Total number of chocolates = 90

a) Probability that she will choose 8 banana and 6 chocolate cakes if she chooses 20 cupcakes at random will be;

(20C₈ × 30C₆ × 40C₆)/90C₂₀

(125970 × 593775 × 3838380)/50980740277700939310

This gives us   0.00563

b) Probability of at least 2 chocolate cupcakes is;

1 - [P(no chocolate) + P(1 chocolate)]

P(no chocolate) = (50C₂₀)/90C₂₀

P(1 chocolate) = (40C₁ × 50C₁₉)/90C₂₀

Thus;

1 - [P(no chocolate) + P(1 chocolate)] = 1 - [(40C₁ × 50C₁₉) + 50C₂₀]/90C₂₀

This gives us;  0.99997522

c) Probability of getting all cupcakes of same kind is;

(40C₂₀ + 20C₂₀ + 30C₂₀)/90C₂₀

⇒ 0.0000000027045

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a) 0.00563

b) 1

c) 0

Step-by-step explanation:

Total = 40+30+20 =90

a) (20C8×30C6×40C6)/90C20

= 0.00563

b) 1 - (no chocolate + 1 chocolate)

1 - [(50C20) + (40C1×50C19)]/90C20

1 - 0.00002478

= 0.9999752187

c) [40C20+20C20+30C20]/90C20

= 0.0000000027045

answer: was that a question or answer?

step-by-step explanation: