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

Answer 1

Answer:  Last Option is correct.

Step-by-step explanation:

Since we have given that

Number of yellow marbles = 8

Number of green marbles = 9

Number of purple marbles = 3

Number of red marbles = 5

So,

Probability that there is atmost one purple marble is given by

$P(X=0)+P(X=1)\\\\=\frac{^{22}C_3}{^25C_3}+\frac{^3C_1\times ^{22}C_2}{^25C_3}\\\\=\frac{22\times 21\times 20}{25\times 24\times 23}+3\times \frac{22\times 21}{25\times 24\times 23}\\\\=0.971$

Hence, Last Option is correct.

Answer 2

Answer:  The correct option is (D) 0.9701.

Step-by-step explanation:  Given that a bag contains eight yellow marbles, nine green marbles, three purple marbles, and five red marbles. Three marbles are randomly chosen from the bag.

We are to find the probability that there is at most one purple marble if three marbles are chosen at random.

Here,

number of yellow marbles = 8,

number of green marbles = 9,

number of purple marbles = 3

and

number of red marbles = 5.

So, total number of marbles = 8 + 9 + 3 + 5 = 25.

If there is at most one purple marble, then there can be 0 or 1 purple marble in the three  randomly chosen marbles.

Therefore, the probability that that there is at most one purple marble if three marbles are chosen at random is given by

$P\\\\\\=P(X=0)+P(X=1)\\\\\\=\dfrac{^{22}C_3}{^{25}C_3}+\dfrac{^3C_1\times ^{22}C_2}{^{25}C_3}\\\\\\=\dfrac{\frac{22!}{3!(22-3)!}}{\frac{25!}{3!(25-3)!}}+\dfrac{3\times \frac{22!}{2!(22-2)!}}{\frac{25!}{3!(25-3)!}}\\\\\\=\dfrac{22!}{3!19!}\times\dfrac{3!22!}{25!}+\dfrac{3\times22!}{2!20!}\times\dfrac{3!22!}{25!}\\\\\\=\dfrac{22\times21\times20}{25\times24\times23}+9\times \dfrac{22\times21}{25\times24\times23}\\\\\\=0.6695+9\times 0.0334\\\\\\=0.9701.$

Thus, the required probability is 0.9701.

Option (D) is correct.