Recall from Example 1 that whenever Suzan sees a bag of marbles, she grabs a handful...

Recall from Example 1 that whenever Suzan sees a bag of marbles, she grabs a handful at random. She has seen a bag containing four red marbles, two green ones, five white ones, and one purple one. She grabs eight of them. Find the probability of the following event, expressing it as a fraction in lowest terms. HINT [See Example 1.]

She has at least one green one.

Solutions

Expert Solution

It is given that whenever Suzan sees a bag of marbles, she grabs a handful at random. The bag contains four red marbles, two green ones, five white ones and one purple one.

We know, = n!/[r!(n-r)!] where, n!=[n(n-1)(n-2).....1]

Thus, the total number of marbles = 4+2+5+1 = 12

It is given that she grabs eight of them.

Thus, the total possible outcomes = = (12*11*10*9)/(4*3*2*1) = 495

We have to find the probability that she has at least one green one.

Now, number of green marbles = 2

Thus, number of marbles without green ones = 12 - 2 = 10

Thus, the number of outcomes that she grabs no green ones = = (10*9)/(2*1) = 45

Let X be a random variable that represents the number of green marbles.

Thus, the probability that she grabs no green ones = P(X = 0) = = 45/495

In the question we have to find the probability that she has at least one green one. Thus, we have to find .

Now,

= 1 - (45/495) = 450/495 = 10/11

Thus, = 10/11

Thus, the probability that she grabs at least one green one = 10/11 .

orchestra answered 1 year ago

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