tisanurnr9c

2023-02-19

A bag contains 5 red marbles and 3 black marbles. Three marbles are drawn one by one without replacement. What is the probability that atleast one of the three marbles drawn be black, if the first marble is red?

i1goqnkb

Beginner2023-02-20Added 5 answers

Let R and B represent red marbles and black marbles respectively.

For a minimum of one of the three marbles chosen to be dark, if the first marble is red, the following three conditions will take place:

(1) Second marble is black and third marble is red i.e., RBR

(2) Second and third marble is black i.e., RBB

(3) Second marble is red and third marble is black i.e., RRB

The necessary likelihood is provided by,

P(atleast one black | first marble is red)

=P(RBR)+P(RBB)+P(RRB)

$=(\frac{5}{8}\times \frac{3}{7}\times \frac{4}{6})+(\frac{5}{8}\times \frac{3}{7}\times \frac{2}{6})+(\frac{5}{8}\times \frac{4}{7}\times \frac{3}{6})$

$=\frac{25}{56}$

For a minimum of one of the three marbles chosen to be dark, if the first marble is red, the following three conditions will take place:

(1) Second marble is black and third marble is red i.e., RBR

(2) Second and third marble is black i.e., RBB

(3) Second marble is red and third marble is black i.e., RRB

The necessary likelihood is provided by,

P(atleast one black | first marble is red)

=P(RBR)+P(RBB)+P(RRB)

$=(\frac{5}{8}\times \frac{3}{7}\times \frac{4}{6})+(\frac{5}{8}\times \frac{3}{7}\times \frac{2}{6})+(\frac{5}{8}\times \frac{4}{7}\times \frac{3}{6})$

$=\frac{25}{56}$