Question

One bag contains three white marbles and five black marbles, and a second bag contains four white marbles and six black marbles. A person draws one marble from each bag. Find the probability that both marbles are black.
A.)55/91
B.)25/72
C.)3/8

Answer 1

Answer:

Probability of getting both balls black = P(BB) = (5/8)×(6/10) = 30/80 = 3/8

Step-by-step explanation:

The given statement is:

One bag contains three white marbles and five black marbles, and a second bag contains four white marbles and six black marbles. A person draws one marble from each bag and we have to find the probability that both the marbles are of black color.

Probability of getting both balls white = P(WW) = (3/8)×(4/10) = 12/80

Probability of getting both balls black = P(BB) = (5/8)×(6/10) = 30/80 = 3/8

Thus the correct option is 3/8 ....

Answer 2

Answer: Option C

$P=\frac{3}{8}$

Step-by-step explanation:

Bag (1)

white marbles: 3

black marbles: 5

total marbles: 8

Probability of taking out a black marble $P(b_1) = \frac{5}{8}$

Bag (2)

white marbles: 4

black marbles: 6

total marbles: 10

Probability of taking out a black marble $P(b_2) = \frac{6}{10}=\frac{3}{5}$

Note that the events are independent. The marble that you take out in the second bag does not depend on the one you took out in the first bag

Then the probability that both marbles are black is:

$P=P(b_1)*P(b_2)$

$(\frac{5}{8})*(\frac{3}{5})$

$P=(\frac{15}{40})$

Simplifying the fraction we get:

$P=(\frac{3}{8})$

The answer is the option C