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

Total marbles \(n = 8 + 9+3 + 5=25\)

Step2: Calculate probability of no purple marble

Number of non - purple marbles \(m=25 - 3=22\)
\(P(X = 0)=\frac{C(22,3)}{C(25,3)}=\frac{\frac{22!}{3!(22 - 3)!}}{\frac{25!}{3!(25 - 3)!}}=\frac{22\times21\times20}{25\times24\times23}=\frac{9240}{13800}\)

Step3: Calculate probability of one purple marble

\(P(X = 1)=\frac{C(3,1)\times C(22,2)}{C(25,3)}=\frac{3\times\frac{22!}{2!(22 - 2)!}}{\frac{25!}{3!(25 - 3)!}}=\frac{3\times231}{2300}=\frac{693}{2300}\)

Step4: Calculate probability of at most one purple marble

\(P(X\leq1)=P(X = 0)+P(X = 1)\)
\(P(X\leq1)=\frac{9240}{13800}+\frac{693}{2300}=\frac{9240+4158}{13800}=\frac{13398}{13800}\approx0.971\)

Answer:

\(0.971\)