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a bag contains eight yellow marbles, nine green marbles, three purple m…

Question

a bag contains eight yellow marbles, nine green marbles, three purple marbles, and five red marbles. three marbles are randomly chosen from the bag. what is the probability that there is at most one purple marble? 0.100 0.301 0.770 0.971

Explanation:

Step1: Calculate total number of marbles

First, find the total number of marbles. Yellow: 8, Green: 9, Purple: 3, Red: 5. So total \( n = 8 + 9 + 3 + 5 = 25 \).

Step2: Define "at most one purple"

"At most one purple" means 0 purple or 1 purple marble. We use combinations. The formula for combinations is \( \binom{n}{k}=\frac{n!}{k!(n - k)!} \).

Step3: Calculate 0 purple marbles

For 0 purple marbles, we choose 3 from non - purple marbles. Non - purple marbles: \( 25 - 3=22 \). So \( \binom{22}{3}=\frac{22!}{3!(22 - 3)!}=\frac{22\times21\times20}{3\times2\times1}=1540 \).

Step4: Calculate 1 purple marble

For 1 purple marble, we choose 1 from 3 purple and 2 from 22 non - purple. \( \binom{3}{1}\times\binom{22}{2}=\ 3\times\frac{22!}{2!(22 - 2)!}=3\times\frac{22\times21}{2\times1}=3\times231 = 693 \).

Step5: Calculate total favorable outcomes

Total favorable outcomes (0 or 1 purple) is \( 1540+693 = 2233 \).

Step6: Calculate total possible outcomes

Total possible outcomes of choosing 3 marbles from 25 is \( \binom{25}{3}=\frac{25!}{3!(25 - 3)!}=\frac{25\times24\times23}{3\times2\times1}=2300 \).

Step7: Calculate probability

Probability \( P=\frac{2233}{2300}\approx0.971 \).

Answer:

0.971