Question

A jar contains 2 green marbles, 4 blue marbles, 3 yellow marbles and 2 black marbles. A marble is chosen at random from the jar and replaced. Then a second marble is chosen at random. What is the probability of the first marble being green and the second marble being yellow?

• A. $\frac{3}{121}$
• B. $\frac{4}{121}$
• C. $\frac{5}{121}$
• D. $\frac{6}{121}$

Answer 1

The correct option is D $\frac{6}{121}$

In the jar, the number of marbles of different colors:

Color of marbles	Number of marbles
Green	$2$
Blue	$4$
Yellow	$3$
Black	$2$

Total number of marbles= $2+4+3+2=11$

Now,
$\text{P(choosing green marble)}=\frac{\text{Number of green marbles}}{\text{Total number of marbles}}$

$\Rightarrow \text{P(choosing green marble)}=\frac{2}{11}$

And,
$\text{P(choosing yellow marble)}=\frac{\text{Number of yellow marbles}}{\text{Total number of marbles}}$

$\Rightarrow \text{P(choosing yellow marble)}=\frac{3}{11}$

$\text{Need to find:}$ The probability of choosing green marble in $1^{\text{st}}$ pick and yellow marble in $2^{\text{nd}}$ pick.

$\therefore \text{P(choosing green and yellow) = P(choosing green)}\times \text{P(choosing yellow)}$

$\Rightarrow \text{P(choosing green and yellow)}=\frac{2}{11}\times \frac{3}{11}$

$\Rightarrow \text{P(choosing green and yellow)}=\frac{6}{121}$

$\therefore$ The Probability of the first marble being green and the second marble being yellow is $\frac{6}{121}$.