Study the following information carefully and answer the given question. The information is related and given in sequence: I. Bag X contains a total o...

Question

Study the following information carefully and answer the given question. The information is related and given in sequence:

I. Bag X contains a total of 30 red and green balls. The probability of selecting one red ball and one green ball from the bag is 4087\frac{40}{87}. The number of red balls in the bag is denoted as A.

II. B number of green balls are taken out, painted red, and placed back into Bag X.

III. The probability of selecting two green balls from Bag X after this change is 729\frac{7}{29}.

IV. Bag Y contains all balls from Bag X. If C green balls are withdrawn from Bag Y and (C5)(C-5) red balls are added, then the probability of selecting two red balls from Bag Y is 25\frac{2}{5}.

Question: Find the probability of selecting (B3)(B-3) green balls from Bag X (considering the initial number of red and green balls in the bag).

Options

A.

3587\frac{35}{87}

B.

1229\frac{12}{29}

C.

4087\frac{40}{87}

D.

3887\frac{38}{87}

E.

3487\frac{34}{87}

probabilityballsbagred and green ballsconditional probability

Solve This Question

Get instant feedback with detailed step-by-step solution

Start Solving →