Question

Bag A contains $9$ purple and $3$ yellow marbles. Bag B contains $9$ blue and $6$ white marbles. What is the probability of picking one yellow marble from bag A and one blue marble from bag B?

• A. $\frac{1}{9}$
• B. $\frac{1}{4}$
• C. $\frac{3}{20}$
• D. $\frac{17}{20}$

Answer 1

The correct option is C $\frac{3}{20}$

The following table shows the number of marbles of each colors in respective bags:

	Purple	Yellow	Blue	White
Bag A	$9$	$3$	$0$	$0$
Bag B	$0$	$0$	$9$	$6$

Total number of marbles in bag A= $9+3=12$
$\text{P(picking yellow marble from bag A)}=\frac{\text{Number of yellow marbles in bag A}}{\text{Total number of marbles in bag A}}$

$\Rightarrow \text{P(picking yellow marble from bag A)}=\frac{3}{12}=\frac{1}{4}$

Next,
Total number of marbles in bag B= $9+6=15$

$\text{P(picking blue marble from bag B)}=\frac{\text{Number of blue marbles in bag B}}{\text{Total number of marbles in bag B}}$

$\Rightarrow \text{P(picking blue marble from bag B)}=\frac{9}{15}$

$\Rightarrow \text{P(picking blue marble from bag B)}=\frac{3\times 3}{3\times 5}=\frac{3}{5}$

$\text{Note:}$ Picking one yellow marble from bag A and one blue marble from bag B are independent compound events.

$\therefore$ Probability of picking one yellow marble from bag A and one blue marble from bag B = $\text{P(picking yellow marble from bag A)}\times \text{P(picking blue marble from bag B)}$

= $\frac{1}{4}\times \frac{3}{5}$

$=\frac{1\times 3}{4\times 5}=\frac{3}{20}$

$\therefore$ The probability of picking one yellow marble from bag A and one blue marble from bag B is $\frac{3}{20}$.